From 8a3d4f0e088bf3bcaea8b567b841da447b2da571 Mon Sep 17 00:00:00 2001 From: huihut Date: Sun, 11 Feb 2018 01:08:43 +0800 Subject: [PATCH] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=E6=9F=A5=E6=89=BE=E7=AE=97?= =?UTF-8?q?=E6=B3=95=E5=92=8CSingleProblem?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- Algorithm/BruteForceStringMatch.h | 14 ++ Algorithm/FileSearch/README.md | 20 ++ Algorithm/FileSearch/input.txt | 5 + Algorithm/FileSearch/search.cpp | 105 ++++++++++ Algorithm/FileSort/README.md | 24 +++ Algorithm/FileSort/input.txt | 1 + Algorithm/FileSort/output.txt | 1 + Algorithm/FileSort/sort.cpp | 194 ++++++++++++++++++ Algorithm/SequentialSearch.h | 9 + .../ChessboardCoverage.cpp | 160 +++++++++++++++ .../ChessboardCoverage.exe | Bin 0 -> 51712 bytes Problems/ChessboardCoverageProblem/README.md | 24 +++ Problems/KnapsackProblem/README.md | 34 +++ Problems/KnapsackProblem/pack.cpp | 165 +++++++++++++++ Problems/KnapsackProblem/pack.exe | Bin 0 -> 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Algorithm/FileSort/input.txt create mode 100644 Algorithm/FileSort/output.txt create mode 100644 Algorithm/FileSort/sort.cpp create mode 100644 Algorithm/SequentialSearch.h create mode 100644 Problems/ChessboardCoverageProblem/ChessboardCoverage.cpp create mode 100644 Problems/ChessboardCoverageProblem/ChessboardCoverage.exe create mode 100644 Problems/ChessboardCoverageProblem/README.md create mode 100644 Problems/KnapsackProblem/README.md create mode 100644 Problems/KnapsackProblem/pack.cpp create mode 100644 Problems/KnapsackProblem/pack.exe create mode 100644 Problems/NeumannNeighborProblem/Formula/Neumann2_3_12.cpp create mode 100644 Problems/NeumannNeighborProblem/Formula/Neumann2_3_12.exe create mode 100644 Problems/NeumannNeighborProblem/Formula/README.md create mode 100644 Problems/NeumannNeighborProblem/README.md create mode 100644 Problems/NeumannNeighborProblem/Recursive/Neumann2_4_12.cpp create mode 100644 Problems/NeumannNeighborProblem/Recursive/Neumann2_4_12.exe create mode 100644 Problems/NeumannNeighborProblem/Recursive/README.md create mode 100644 Problems/RoundRobinProblem/MatchTable.cpp create mode 100644 Problems/RoundRobinProblem/MatchTable.exe create mode 100644 Problems/RoundRobinProblem/README.md create mode 100644 Problems/TubingProblem/README.md create mode 100644 Problems/TubingProblem/Tubing.cpp create mode 100644 Problems/TubingProblem/Tubing.exe diff --git a/Algorithm/BruteForceStringMatch.h b/Algorithm/BruteForceStringMatch.h new file mode 100644 index 0000000..5354509 --- /dev/null +++ b/Algorithm/BruteForceStringMatch.h @@ -0,0 +1,14 @@ +// 蛮力字符串匹配 +int BruteForceStringMatch(vector& Total, vector& Part) +{ + int i, j; + for (i = 0; i < Total.size() - Part.size(); ++i) { + j = 0; + while (j < Part.size() && Part[j] == Total[i + j]) { + ++j; + if (j == Part.size()) + return i; + } + } + return -1; +} \ No newline at end of file diff --git a/Algorithm/FileSearch/README.md b/Algorithm/FileSearch/README.md new file mode 100644 index 0000000..c150b8a --- /dev/null +++ b/Algorithm/FileSearch/README.md @@ -0,0 +1,20 @@ + +## 文件（文本）查找 + +### 代码 + +[文件（文本）查找代码](search.cpp) + +### 功能说明 + +本程序实现对英文文本中关键字的查找 + +返回关键字出现的位置（第几个词） + +### 代码简述 + +从`output.txt`文件读入数据（英文文本）到 `vector` 中存储 + +通过用户输入关键字（`keyword`）进行查找 + +输出查找到的关键字出现的位置（第几个词） \ No newline at end of file diff --git a/Algorithm/FileSearch/input.txt b/Algorithm/FileSearch/input.txt new file mode 100644 index 0000000..2c103a3 --- /dev/null +++ b/Algorithm/FileSearch/input.txt @@ -0,0 +1,5 @@ +Just arrived in Kathmandu tonight and needed to eat, +we checked Trip advisor and came here because it was close to hotel. +Certainly deserves its number 1 rating. +Food is so tasty and the atmosphere is very relaxed. +I would definitely recommend. \ No newline at end of file diff --git a/Algorithm/FileSearch/search.cpp b/Algorithm/FileSearch/search.cpp new file mode 100644 index 0000000..19e2e74 --- /dev/null +++ b/Algorithm/FileSearch/search.cpp @@ -0,0 +1,105 @@ +#include +#include +#include +#include + +using namespace std; + +#define OK 0 +#define ERROR -1 +#define INPUTNAME "input.txt" + +// 查找 +int search(vector& v_data, string keyword) { + int i = 0; + + // 遍历每个元素直到找到关键字 + for (; i < v_data.size(); i++) { + + // 比较关键字与每个词是否相等 + if (0 == v_data[i].compare(keyword)){ + + // 相等则返回关键字的位序 + return i; + } + } + return ERROR; +} + + +// 文件读取 +int FileRead(vector& v_data) { + + ifstream f_in(INPUTNAME); + + //判断读取失败 + if (!f_in) { + cout << "文件读取失败！" << endl; + system("pause"); + return ERROR; + } + + string word; + + // 文件逐个读取 + while (!f_in.eof()) { + f_in >> word; + v_data.push_back(word); + } + + // 关闭文件 + f_in.close(); + + return OK; +} + +// 界面 +void Interface(string& keyword) { + cout << "-------------------- 文件查找 --------------------" << endl; + cout << "【说明】：本程序实现对文件内容的查找" << endl; + cout << "即从input.txt读入，查找关键字的首次出现位置" << endl; + cout << "--------------------------------------------------" << endl; + cout << "请输入关键字：" << endl; + cout << "--------------------------------------------------" << endl; + cin >> keyword; + cout << "--------------------------------------------------" << endl; +} + +int main() { + + vector v_data; + + // 文件读取 + if (ERROR == FileRead(v_data)) + return ERROR; + + int index = -1; + string keyword; + + // 界面 + Interface(keyword); + + // 合法性检测 + if (keyword.empty()) { + cout << "请输入合法的关键字！" << endl; + system("pause"); + return ERROR; + } + + // 查找 + index = search(v_data, keyword); + + //未找到输出 + if (ERROR == index) { + cout << "未找到此关键字！" << endl; + system("pause"); + return ERROR; + } + + //输出找到的关键字索引 + cout << "此关键字位于第 " << index + 1 << " 个词的位置！" << endl; + + system("pause"); + + return OK; +} diff --git a/Algorithm/FileSort/README.md b/Algorithm/FileSort/README.md new file mode 100644 index 0000000..2704e10 --- /dev/null +++ b/Algorithm/FileSort/README.md @@ -0,0 +1,24 @@ + +## 文件（文本）排序 + +### 代码 + +[文件（文本）排序代码](sort.cpp) + +### 功能说明 + +本程序实现选择排序和冒泡排序两个排序算法 + +并且有从小到大和从大到小两种排序方式 + +用户可进行选择需要的方式 + +### 代码简述 + +从 `output.txt` 文件读入数据（数字）到 `vector` 中存储 + +通过用户输入的排序算法（`i_algorithm`）和排序方式（`i_mode`） + +选择对于的选择排序（`SelectSort()`）或者冒泡排序（`BubbleSort()`）进行排序 + +排序后输出 `vector` 到 `output.txt` \ No newline at end of file diff --git a/Algorithm/FileSort/input.txt b/Algorithm/FileSort/input.txt new file mode 100644 index 0000000..343ab36 --- /dev/null +++ b/Algorithm/FileSort/input.txt @@ -0,0 +1 @@ +33 22 66 99 11 68 39 89 107 749 20 6 \ No newline at end of file diff --git a/Algorithm/FileSort/output.txt b/Algorithm/FileSort/output.txt new file mode 100644 index 0000000..a136e47 --- /dev/null +++ b/Algorithm/FileSort/output.txt @@ -0,0 +1 @@ +6 11 20 22 33 39 66 68 89 99 107 749 \ No newline at end of file diff --git a/Algorithm/FileSort/sort.cpp b/Algorithm/FileSort/sort.cpp new file mode 100644 index 0000000..94beeca --- /dev/null +++ b/Algorithm/FileSort/sort.cpp @@ -0,0 +1,194 @@ +#include +#include +#include + +using namespace std; + +#define OK 0 +#define ERROR -1 +#define INPUTNAME "input.txt" +#define OUTPUTNAME "output.txt" + +// 选择排序 +int SelectSort(vector& v_data, int b_mode) { + + size_t i_num = v_data.size(), temp = 0; + + // 从小到大排 + if (0 == b_mode) { + size_t min; + for (size_t i = 0; i < i_num - 1; i++) { + min = i; + for (size_t j = i + 1; j < i_num; j++) + if (v_data[min] > v_data[j]) + min = j; + if (min != i) { + temp = v_data[min]; + v_data[min] = v_data[i]; + v_data[i] = temp; + } + } + } + + // 从大到小排 + else { + size_t max; + for (size_t i = 0; i < i_num - 1; i++) { + max = i; + for (size_t j = i + 1; j < i_num; j++) + if (v_data[max] < v_data[j]) + max = j; + if (max != i) { + temp = v_data[max]; + v_data[max] = v_data[i]; + v_data[i] = temp; + } + } + } + + return OK; +} + +// 冒泡排序 +int BubbleSort(vector& v_data, int b_mode) { + + size_t num = v_data.size(), temp = 0; + + // 从小到大排 + if (0 == b_mode) { + for (size_t i = 0; i < num - 1; i++) { + for (size_t j = 0; j < num - i - 1; j++) { + if (v_data[j] > v_data[j + 1]) { + temp = v_data[j]; + v_data[j] = v_data[j + 1]; + v_data[j + 1] = temp; + } + } + } + } + + // 从大到小排 + else { + for (size_t i = 0; i < num - 1; i++) { + for (size_t j = 0; j < num - i - 1; j++) { + if (v_data[j] < v_data[j + 1]) { + temp = v_data[j]; + v_data[j] = v_data[j + 1]; + v_data[j + 1] = temp; + } + } + } + } + + return OK; +} + +// 文件读取 +int FileRead(vector& v_data) { + + ifstream f_in(INPUTNAME); + + //判断读取失败 + if (!f_in) { + cout << "文件读取失败！" << endl; + system("pause"); + return ERROR; + } + + int i_temp; + + // 文件逐个读取 + while (!f_in.eof()) { + f_in >> i_temp; + v_data.push_back(i_temp); + } + + // 关闭文件 + f_in.close(); + + return OK; +} + +// 文件写入 +int FileWrite(vector& v_data) { + + ofstream f_out(OUTPUTNAME); + + // 判断读取失败 + if (!f_out) { + cout << "文件写入失败！" << endl; + return ERROR; + } + + // 文件逐个写入 + for (int i = 0; i < v_data.size(); i++) + f_out << v_data[i] << " "; + + f_out.close(); + return OK; +} + +// 界面 +void Interface(int& i_algorithm, int& i_mode) { + cout << "-------------------- 文件排序 --------------------" << endl; + cout << "【说明】：本程序实现对文件内容的排序" << endl; + cout << "即从input.txt读入，排序后写入到output.txt" << endl; + cout << "--------------------------------------------------" << endl; + cout << "请选择排序算法：" << endl; + cout << "【0】选择排序" << endl; + cout << "【1】冒泡排序" << endl; + cout << "--------------------------------------------------" << endl; + cin >> i_algorithm; + cout << "--------------------------------------------------" << endl; + cout << "请选择排序方式：" << endl; + cout << "【0】从小到大" << endl; + cout << "【1】从大到小" << endl; + cout << "--------------------------------------------------" << endl; + cin >> i_mode; + cout << "--------------------------------------------------" << endl; +} + +int main() { + + vector v_data; + + // 文件读取 + if (ERROR == FileRead(v_data)) + return ERROR; + + int i_algorithm, i_mode; + + // 界面 + Interface(i_algorithm, i_mode); + + // 排序算法选择检测 + if (0 != i_algorithm && 1 != i_algorithm) { + cout << "排序算法选择错误！" << endl; + system("pause"); + return ERROR; + } + + // 排序方式选择检测 + if (0 != i_mode && 1 != i_mode) { + cout << "排序方式选择错误！" << endl; + system("pause"); + return ERROR; + } + + // 排序 + if (i_algorithm) + BubbleSort(v_data, i_mode); + else + SelectSort(v_data, i_mode); + + // 文件写入 + if (ERROR == FileWrite(v_data)) + return ERROR; + + cout << "排序完成，数据已写入：" << OUTPUTNAME << endl; + cout << "--------------------------------------------------" << endl; + + system("pause"); + + return OK; +} diff --git a/Algorithm/SequentialSearch.h b/Algorithm/SequentialSearch.h new file mode 100644 index 0000000..bef7ced --- /dev/null +++ b/Algorithm/SequentialSearch.h @@ -0,0 +1,9 @@ +// 顺序查找 +int SequentialSearch(vector& v, int k) { + int i = 0; + for (; i < v.size(); ++i) + if (v[i] == k) + return i; + if (i == v.size()) + return -1; +} \ No newline at end of file diff --git a/Problems/ChessboardCoverageProblem/ChessboardCoverage.cpp b/Problems/ChessboardCoverageProblem/ChessboardCoverage.cpp new file mode 100644 index 0000000..57cf1bd --- /dev/null +++ b/Problems/ChessboardCoverageProblem/ChessboardCoverage.cpp @@ -0,0 +1,160 @@ +#include +#include +#include + +using namespace std; + +int num_Now = 0; // 记录L型骨牌编号 +int **board = NULL; // 棋盘指针 + + // 函数声明 +void ChessBoard(int num_BoardTopLeftRow, int num_BoardTopLeftColumn, int num_SpecialRow, int num_SpecialColumn, int boardSize); + +int main() { + + int num_BoardTopLeftRow = 0, // 棋盘左上角的行号 + num_BoardTopLeftColumn = 0, // 棋盘左上角的列号 + num_SpecialRow = 0, // 特殊方格所在的行号 + num_SpecialColumn = 0, // 特殊方格所在的列号 + boardSize = 0, // 棋盘大小 + k = 0; // 构成的(2^k)*(2^k)个方格的棋盘 + + // 用户界面 + cout << "---------------- 棋盘覆盖问题 ----------------" << endl; + cout << "请输入k(k>=0)，构成(2^k)*(2^k)个方格的棋盘" << endl; + + // 输入k值 + cin >> k; + + // 判断输入数据合法性，包括检查输入是否为数字，k值是否大于0 + if (cin.fail() || k < 0) + { + cout << "输入k错误！" << endl; + system("pause"); + return 0; + } + + // 计算棋盘大小 + boardSize = pow(2, k); + + cout << "请输入特殊方格所在的行号和列号（从0开始，用空格隔开）" << endl; + + // 输入特殊方格所在的行号和列号 + cin >> num_SpecialRow >> num_SpecialColumn; + + // 判断输入数据合法性，包括检查输入是否为数字，特殊方格行号列号是否大于0，特殊方格行号列号是否不大于棋盘大小 + if (cin.fail() || num_SpecialRow < 0 || num_SpecialColumn < 0 || num_SpecialRow >= boardSize || num_SpecialColumn >= boardSize) + { + cout << "输入行号或列号错误！" << endl; + system("pause"); + return 0; + } + + // 分配棋盘空间 + board = new int *[boardSize]; + for (auto i = 0; i < boardSize; i++) + { + board[i] = new int[boardSize]; + } + + // 为特殊方格赋初值0 + board[num_SpecialRow][num_SpecialColumn] = 0; + + //执行棋盘覆盖函数 + ChessBoard(num_BoardTopLeftRow, num_BoardTopLeftColumn, num_SpecialRow, num_SpecialColumn, boardSize); + + // 显示输出 + cout << "------------------------------------------------" << endl; + for (auto i = 0; i < boardSize; i++) + { + for (auto j = 0; j < boardSize; j++) + { + cout << board[i][j] << "\t"; + } + cout << endl; + } + cout << "------------------------------------------------" << endl; + + // 暂停查看结果 + system("pause"); + + // 释放内存 + for (int i = 0; i <= boardSize; i++) + delete[] board[i]; + delete[] board; + + // 指针置空 + board = NULL; + + return 0; +} + +// 棋盘覆盖函数 +void ChessBoard(int num_BoardTopLeftRow, int num_BoardTopLeftColumn, int num_SpecialRow, int num_SpecialColumn, int boardSize) +{ + // 棋盘大小为1则直接返回 + if (boardSize == 1) return; + + int num = ++num_Now, // L型骨牌编号 + size = boardSize / 2; // 分割棋盘，行列各一分为二 + + // 覆盖左上角子棋盘 + if (num_SpecialRow < num_BoardTopLeftRow + size && num_SpecialColumn < num_BoardTopLeftColumn + size) + { + // 递归覆盖含有特殊方格的子棋盘 + ChessBoard(num_BoardTopLeftRow, num_BoardTopLeftColumn, num_SpecialRow, num_SpecialColumn, size); + } + else + { + // 用编号为num的L型骨牌覆盖右下角 + board[num_BoardTopLeftRow + size - 1][num_BoardTopLeftColumn + size - 1] = num; + + // 递归覆盖其余棋盘 + ChessBoard(num_BoardTopLeftRow, num_BoardTopLeftColumn, num_BoardTopLeftRow + size - 1, num_BoardTopLeftColumn + size - 1, size); + } + + // 覆盖右上角子棋盘 + if (num_SpecialRow < num_BoardTopLeftRow + size && num_SpecialColumn >= num_BoardTopLeftColumn + size) + { + // 递归覆盖含有特殊方格的子棋盘 + ChessBoard(num_BoardTopLeftRow, num_BoardTopLeftColumn + size, num_SpecialRow, num_SpecialColumn, size); + } + else + { + // 用编号为num的L型骨牌覆盖左下角 + board[num_BoardTopLeftRow + size - 1][num_BoardTopLeftColumn + size] = num; + + // 递归覆盖其余棋盘 + ChessBoard(num_BoardTopLeftRow, num_BoardTopLeftColumn + size, num_BoardTopLeftRow + size - 1, num_BoardTopLeftColumn + size, size); + } + + // 覆盖左下角子棋盘 + if (num_SpecialRow >= num_BoardTopLeftRow + size && num_SpecialColumn < num_BoardTopLeftColumn + size) + { + // 递归覆盖含有特殊方格的子棋盘 + ChessBoard(num_BoardTopLeftRow + size, num_BoardTopLeftColumn, num_SpecialRow, num_SpecialColumn, size); + } + else + { + // 用编号为num的L型骨牌覆盖右上角 + board[num_BoardTopLeftRow + size][num_BoardTopLeftColumn + size - 1] = num; + + // 递归覆盖其余棋盘 + ChessBoard(num_BoardTopLeftRow + size, num_BoardTopLeftColumn, num_BoardTopLeftRow + size, num_BoardTopLeftColumn + size - 1, size); + } + + // 覆盖右下角子棋盘 + if (num_SpecialRow >= num_BoardTopLeftRow + size && num_SpecialColumn >= num_BoardTopLeftColumn + size) + { + // 递归覆盖含有特殊方格的子棋盘 + ChessBoard(num_BoardTopLeftRow + size, num_BoardTopLeftColumn + size, num_SpecialRow, num_SpecialColumn, size); + } + else + { + // 用编号为num的L型骨牌覆盖左上角 + board[num_BoardTopLeftRow + size][num_BoardTopLeftColumn + size] = num; + + // 递归覆盖其余棋盘 + ChessBoard(num_BoardTopLeftRow + size, num_BoardTopLeftColumn + size, num_BoardTopLeftRow + size, num_BoardTopLeftColumn + size, size); + } +} diff --git a/Problems/ChessboardCoverageProblem/ChessboardCoverage.exe b/Problems/ChessboardCoverageProblem/ChessboardCoverage.exe new file mode 100644 index 0000000000000000000000000000000000000000..0ebc7c2d9664e1720fe1290a25eb673a47b46d41 GIT binary patch literal 51712 zcmeHw4M3Dt*8d$CbTkk&C@QL{AT2)v2ASbw1e7r>u>l3F%yEQ)0r_%1{Ai`bk%$16?xXNgS z1M<`T(_3V$9ITw~nr1VtvMse*8S}_B4B``vwR1e=?N|vwIv|}4;#0}kl4+JAqs_?J zuVRq@0PwEGeGP)&l@7&ClhBxEG%^~1N3@>wLqaqiInzod#HYq10kW_OxIc~Hccl}( 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b/Problems/ChessboardCoverageProblem/README.md new file mode 100644 index 0000000..01d6755 --- /dev/null +++ b/Problems/ChessboardCoverageProblem/README.md @@ -0,0 +1,24 @@ + +## 棋盘覆盖问题 + +### 代码 + +[棋盘覆盖问题代码](ChessboardCoverage.cpp) + +### 问题说明 + +在一个2^k * 2^k个方格组成的棋盘中，恰有一个方格与其它方格不同，称该方格为一特殊方格。 + +棋盘覆盖问题就是要用图示的4种不同形态的L型骨牌覆盖给定棋盘上除特殊方格之外的所有方格，且任何2个L型骨牌不得重叠覆盖。 + +![](http://blog.chinaunix.net/attachment/201303/1/26548237_1362125215RWwI.png) + +### 功能说明 + +本程序用分治法的思想解决了棋盘覆盖问题，显示输出 + +### 代码简述 + +用户输入数据，程序输入检测，动态分配空间，调用棋盘覆盖函数，把计算结果存储到board（二维数组指针），显示输出。 + +其中棋盘覆盖函数用分治的思想把棋盘分成四份，递归求解。 \ No newline at end of file diff --git a/Problems/KnapsackProblem/README.md b/Problems/KnapsackProblem/README.md new file mode 100644 index 0000000..08ef1df --- /dev/null +++ b/Problems/KnapsackProblem/README.md @@ -0,0 +1,34 @@ + +## 背包问题 + +### 代码 + +[背包问题代码](pack.cpp) + +### 问题说明 + +有N件物品和一个容量为V的背包。 + +第i件物品的重量是w[i]，价值是v[i]。 + +求解将哪些物品装入背包可使这些物品的重量总和不超过背包容量， + +且价值总和最大。 + +### 功能说明 + +本程序用动态规划的思想解决了背包问题，并用了两种算法： +迭代法、递归法。在迭代法中实现了打印背包问题的表格。 + +### 代码简述 + +通过用户输入数据，程序输入检测，动态分配空间，选择算法， +用动态规划的思想求解背包问题。 + +#### 迭代法： +通过遍历n行W列，迭代每行每列的值，并把最优解放到 +n行（在数组中为第n+1行）W列（在数组中为第W+1列）中。 + +#### 递归法： +通过每次返回前i个物品和承重为j的最优解， +递归计算总背包问题的最优解。 diff --git a/Problems/KnapsackProblem/pack.cpp b/Problems/KnapsackProblem/pack.cpp new file mode 100644 index 0000000..7404516 --- /dev/null +++ b/Problems/KnapsackProblem/pack.cpp @@ -0,0 +1,165 @@ +#include +#include +using namespace std; + +int *w = NULL; // 存储每件物品重量的数组指针 +int *v = NULL; // 存储每件物品价值的数组指针 +int **T = NULL; // 存储背包问题表格的数组指针 +int n; // 物品个数n +int W; // 背包总承重W + +// 返回两个值的最大值 +int max(int a, int b) +{ + return (a > b) ? a : b; +} + +// 迭代法，能显示背包问题的表格 +void packIterative() +{ + // 循环遍历n行 + for (int i = 1; i <= n; ++i) + { + // 循环遍历W列 + for (int j = 1; j <= W; ++j) + { + //第i个物品能装下，则比较包括第i个物品和不包括第i个物品，取其最大值 + if (w[i] <= j) + T[i][j] = max(v[i] + T[i - 1][j - w[i]], T[i - 1][j]); + + // 第i个物品不能装下，则递归装i-1个 + else + T[i][j] = T[i - 1][j]; + } + } +} + +// 递归法，不支持显示背包问题的表格 +int packRecursive(int i, int j, int *w, int *v) +{ + // 结束条件（初始条件），i或者j为0时最大总价值为0 + if (i == 0 || j == 0) + return 0; + + // 第i个物品不能装下，则递归装i-1个 + if (w[i] > j) + return packRecursive(i - 1, j, w, v); + + //第i个物品能装下，则比较包括第i个物品和不包括第i个物品，取其最大值 + else + return max(v[i] + packRecursive(i - 1, j - w[i], w, v), packRecursive(i - 1, j, w, v)); +} + +// 打印背包问题的表格 +void printT(int n, int W) +{ + // 打印n行 + for (auto i = 0; i <= n; i++) + { + // 打印行数 + cout << i << ":\t"; + + // 打印W列 + for (int w = 0; w <= W; w++) + { + cout << T[i][w] << "\t"; + } + + // 换行 + cout << endl; + } +} + +int main() +{ + + cout << "---------------- 背包问题 ----------------" << endl; + cout << "请输入物品数 n (n>=0) " << endl; + + // 输入背包数 + cin >> n; + + if (cin.fail() || n < 0) + { + cout << "输入n错误！" << endl; + system("pause"); + return 0; + } + + cout << "请输入背包承重量 W (W>=0) " << endl; + + // 输入背包承重量 + cin >> W; + + if (cin.fail() || W < 0) + { + cout << "输入W错误！" << endl; + system("pause"); + return 0; + } + + // 分配空间 + // 对w和v分配n+1大小 + w = new int[n+1]; + v = new int[n+1]; + + // 对T分配n+1行，并初始化为0 + T = new int *[n + 1](); + // 对T分配W+1列，并初始化为0 + for (auto i = 0; i <= n; i++) + { + T[i] = new int[W + 1](); + } + + // 输入背包的重量和价值 + for (auto i = 1; i <= n; i++) + { + cout << "请输入第 " << i << " 个背包的重量和价值（用空格隔开）" << endl; + cin >> w[i] >> v[i]; + if (cin.fail() || w[i] < 0 || v[i] < 0) + { + cout << "输入错误！" << endl; + system("pause"); + return 0; + } + } + + cout << "------------------------------------------------" << endl; + cout << "请选择算法：" << endl; + cout << "【1】迭代法" << endl; + cout << "【2】递归法" << endl; + cout << "------------------------------------------------" << endl; + + int choose; + + // 输入算法的选择 + cin >> choose; + switch (choose) + { + case 1: + { + // 迭代法，能显示背包问题的表格 + packIterative(); + cout << "能装下物品的最大价值为 " << T[n][W] << endl; + cout << "------------------------------------------------" << endl; + printT(n, W); + break; + } + case 2: + { + // 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阶冯诺依曼邻居的元胞数\n"); + printf("----------------------------\n"); + printf("请输入n\n"); + scanf("%d", &n); + + //用通项公式求解 + a = Neumann2_3_12(n); + + printf("------------通项法-------------\n"); + printf(" %d 阶冯诺依曼邻居的元胞数为 %d\n", n, a); + + getchar(); + getchar(); + return 0; +} diff --git a/Problems/NeumannNeighborProblem/Formula/Neumann2_3_12.exe b/Problems/NeumannNeighborProblem/Formula/Neumann2_3_12.exe new file mode 100644 index 0000000000000000000000000000000000000000..6ed7dfe9539d425b018c2738f020ec5257782857 GIT binary patch literal 37888 zcmeHw4`9>Pwf{|Tc$_|ZAw!5kNhd%0I8vS zgs<7$9{SYj9thifKKI7lYdcr5)9O0&`<(kt(xepsY`*=r zy$0_2-aqG_d(OG%p6~sh`%PoM?Oqno7?UFfgN(J|N}qfVKN}39`QodextQ&paP*os zS@F?pmN=YlRg$M}P z_x$Y9qHxNcZ7%v2%1`y}TJUXt{n3K&@#_y4wBTA|cJlUnezvRVWt6A({b)e|*Q@&O zD5C4sz8$<=>a21Q%?n9jF)>ywo4_{z@R_AyoNg8;yGV91V;0mDO0~Bn%NR>Wo)T$= zu5pZwqar@xE9(ayDG{G|mfyk=c^?3Nja)d+vpZ#c7pUynZpK1#je@v?v2Kosw*JKcqyr1CBluJ?wqmxc z+UBt_R(=iIR{-x?T$dn4eexm3Yzd8RNkGHBz=IB1a}*M?`HZ#Cc1nnkPDBM{VM<)r 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+![](http://ojlsgreog.bkt.clouddn.com/NeumannNeighborProblem.jpg) + +### 功能说明 + +本程序使用通项公式求解。 + +### 代码简述 + +若设第n次生成的方格数是a(n)，则： + +a(1) = a(0) + 4 * 1 +a(2) = a(1) + 4 * 2 +a(3) = a(2) + 4 * 3 +... +a(n) = a(n-1) + 4 * n + +化简可得： + +a(n) - a(1) = 4 * (n + (n-1) + ... + 2 ) + +即： + +a(n) = 2 * n*n + 2 * n + 1 + +则可得出a(n)的通项公式，即可用通项公式直接求解。 + +在程序中用Neumann2_3_12函数返回a(n)的值。 \ No newline at end of file diff --git a/Problems/NeumannNeighborProblem/README.md b/Problems/NeumannNeighborProblem/README.md new file mode 100644 index 0000000..c2e76a4 --- /dev/null +++ b/Problems/NeumannNeighborProblem/README.md @@ -0,0 +1,13 @@ + +## 冯诺依曼邻居问题 + +### 问题说明 + +某算法从一个1×1的方格开始，每次都会在上次图形的周围再加上一圈方格，在第n次的时候要生成多少个方格？下图给出了n = 0，1，2是的结果。 + +![](http://ojlsgreog.bkt.clouddn.com/NeumannNeighborProblem.jpg) + +### 解法 + +* [通项公式解法](Formula) +* [递推关系解法](Recursive) \ No newline at end of file diff --git a/Problems/NeumannNeighborProblem/Recursive/Neumann2_4_12.cpp b/Problems/NeumannNeighborProblem/Recursive/Neumann2_4_12.cpp new file mode 100644 index 0000000..77b2a92 --- /dev/null +++ b/Problems/NeumannNeighborProblem/Recursive/Neumann2_4_12.cpp @@ -0,0 +1,36 @@ +#include + +//递归法 +int Neumann2_4_12(int n) { + + //由图可知第0次有1个方格 + if (n == 0) return 1; + + //递推关系的求解请查看说明文档 + return Neumann2_4_12(n - 1) + 4 * n; +} + +int main() { + int n = 0, a = 0; + + printf("------冯诺依曼邻居问题------\n"); + printf("已知：\n"); + printf(" 0 阶冯诺依曼邻居的元胞数为 1 \n"); + printf(" 1 阶冯诺依曼邻居的元胞数为 5 \n"); + printf(" 2 阶冯诺依曼邻居的元胞数为 13 \n"); + printf("求：\n"); + printf(" n 阶冯诺依曼邻居的元胞数\n"); + printf("----------------------------\n"); + printf("请输入n\n"); + scanf("%d", &n); + + //建立递推关系，使用递归求解 + a = Neumann2_4_12(n); + + printf("------------通项法-------------\n"); + printf(" %d 阶冯诺依曼邻居的元胞数为 %d\3n", n, a); + + getchar(); + getchar(); + return 0; +} diff --git a/Problems/NeumannNeighborProblem/Recursive/Neumann2_4_12.exe b/Problems/NeumannNeighborProblem/Recursive/Neumann2_4_12.exe new 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z&b~XlC3|1?lAOnK7HMwPe5jeIy;Pg2E!N(m^=Kc`wrLM&-_?GojnO6R*6V%@DelnU zt^cw9Dg6ujclGb<;|)oMD-G8g<``}H@yw{rI99L)JRN2XC~F47cgiZzQh zOEs06YRxLm8qJ-WZ)pOWt(qUgqW`7&AI)zx9h$?MS2e%Wyn}JRKzq6NTJ2-nUD_wK z&uI5)JG3urU(^0hdrI4_J)`|ZE7OhBP0*$2uF_4_rR%ghi>_4nw60aRUw26Nitf1X zE!}C|pLBh?&vY^Rarz1ROZ8Xjr|Z-8v-Np;w|M4aJ7VhNXr|L$#s7&}dj^xXZBF@EzFgLBlq~PYjP6er0GeJZES#95fs;yk +#include + +using namespace std; + +// 循环赛日程安排函数声明 +void MatchTable(int k, int n, int **table); + +int main() +{ + int n = 0, k = 0; + + // 用户界面 + cout << "---------------- 循环赛日程安排问题 ----------------" << endl; + cout << "请输入k(k>=0)，构成 n=(2^k) 个选手的循环赛" << endl; + + // 输入k值 + cin >> k; + + // 判断输入数据合法性，包括检查输入是否为数字，k值是否大于0 + if (cin.fail() || k < 0) + { + cout << "输入k错误！" << endl; + system("pause"); + return 0; + } + + // 计算比赛日程表大小 + n = pow(2, k); + + // 分配日程表空间 + int **table = new int *[n + 1]; + for (int i = 0; i <= n; i++) + { + table[i] = new int[n + 1]; + } + + // 进行循环赛日程安排，生成日程表 + MatchTable(k, n, table); + + // 显示输出 + cout << "------------------------------------------------" << endl; + for (int i = 1; i <= n; i++) + { + for (int j = 1; j <= n; j++) + { + cout << table[i][j] << "\t"; + } + cout << endl; + } + cout << "------------------------------------------------" << endl; + + // 暂停查看结果 + system("pause"); + + // 释放内存 + for (int i = 0; i <= n; i++) + delete[] table[i]; + delete[] table; + + // 指针置空 + table = NULL; + + return 0; +} + +// 进行循环赛日程安排，生成日程表 +void MatchTable(int k, int n, int **table) +{ + // 设置日程表第一行的值 + for (int i = 1; i <= n; i++) + table[1][i] = i; + + // 每次填充的起始填充位置 + int begin = 1; + + // 用分治法分separate份，循环求解 + for (int separate = 1; separate <= k; separate++) + { + // 日程表进行划分 + n /= 2; + + // flag为每一小份的列的标记 + for (int flag = 1; flag <= n; flag++) + { + // 操作行 + for (int i = begin + 1; i <= 2 * begin; i++) + { + // 操作列 + for (int j = begin + 1; j <= 2 * begin; j++) + { + // 把左上角的值赋给右下角 + table[i][j + (flag - 1) * begin * 2] = table[i - begin][j + (flag - 1) * begin * 2 - begin]; + // 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zkEI?@6;uC`+MD`KD$|V8C^Zu_6E&A;;x$)k5;dusOwBCKT+IT_BFzn&WtuWgrN*XN zqq#-1NwYbsaw>o>TY$9x>pUFl!+<1sdG`Aw$yBGE_(Gc=zjzH$~NszZG(1?wn=*o zeWg#UOp8gIm=>27pO%@nC~Yy?#gVoy?VdEPj_s83xnpkPlH}UtHOcoRZ$nwV$&1vB z)eiL<^vwIzyVMQpCY0N*KBgAc{c1%@T*~B>#FWgG?3COTLrN*VoHj~z%T6jy+MLvY W&jP~u`zk)03SA)$gfws(4g5c{*kcy} literal 0 HcmV?d00001 diff --git a/Problems/RoundRobinProblem/README.md b/Problems/RoundRobinProblem/README.md new file mode 100644 index 0000000..af58486 --- /dev/null +++ b/Problems/RoundRobinProblem/README.md @@ -0,0 +1,31 @@ + +## 循环赛日程安排问题 + +### 代码 + +[循环赛日程安排问题代码](MatchTable.cpp) + +### 问题说明 + +设有n=2k个选手要进行网球循环赛， +要求设计一个满足以下要求的比赛日程表： + +（1）每个选手必须与其他n-1个选手各赛一次； +（2）每个选手一天只能赛一次。 + +按此要求，可将比赛日程表设计成一个 n 行n-1列的二维表， +其中，第 i 行第 j 列表示和第 i 个选手在第 j 天比赛的选手。 + +### 功能说明 + +本程序运用分治的思想，实现了循环赛日程安排问题的求解， +生成日程表，输出。 + +### 代码简述 + +通过用户输入数据，程序输入检测，动态分配空间， +调用生成日程表函数，显示输出。 + +其中，生成日程表函数运用分治的思想，分成separate份， +先安排第一行（第一份），然后每一份填充，最终求解完毕， +生成日程表。 \ No newline at end of file diff --git a/Problems/TubingProblem/README.md b/Problems/TubingProblem/README.md new file mode 100644 index 0000000..0828556 --- /dev/null +++ b/Problems/TubingProblem/README.md @@ -0,0 +1,31 @@ + +## 输油管道问题 + +### 代码 + +[输油管道问题代码](Tubing.cpp) + +### 问题说明 + +某石油公司计划建造一条由东向西的主输油管道。 +该管道要穿过一个有n 口油井的油田。 +从每口油井都要有一条输油管道沿最短路经(或南或北)与主管道相连。 +如果给定n口油井的位置,即它们的x 坐标（东西向）和y 坐标（南北向）, +应如何确定主管道的最优位置, +即使各油井到主管道之间的输油管道长度总和最小的位置? + +### 功能说明 + +本程序用排序求中值的方法求解输油管道问题。 + +### 代码简述 + +通过用户输入数据（只输入油井数n、每个油井的y坐标）， +程序输入检测，动态分配空间，排序（使用快速排序）， +求出中间值，输出。 + +输出有以下两种情况： + +1. 当n为奇数，则最优位置为y数组的第n/2个油井的y坐标 + +2. 当n为偶数，则最优位置为y数组的中间两个油井的y坐标的区间 \ No newline at end of file diff --git a/Problems/TubingProblem/Tubing.cpp b/Problems/TubingProblem/Tubing.cpp new file mode 100644 index 0000000..59f4e02 --- /dev/null +++ b/Problems/TubingProblem/Tubing.cpp @@ -0,0 +1,116 @@ +#include +using namespace std; + +// 油井y坐标指针 +float * y = NULL; + +// 快速排序 +void quick_sort(int low, int high) +{ + if (low >= high) // 结束标志 + return; + int first = low; // 低位下标 + int last = high; // 高位下标 + float key = y[first]; // 设第一个为基准 + + while (first < last) + { + // 将比第一个小的移到前面 + while (first < last && y[last] >= key) + last--; + if (first < last) + y[first++] = y[last]; + + // 将比第一个大的移到后面 + while (first < last && y[first] <= key) + first++; + if (first < last) + y[last--] = y[first]; + } + // 基准置位 + y[first] = key; + // 前半递归 + quick_sort(low, first - 1); + // 后半递归 + quick_sort(first + 1, high); +} + +int main() +{ + int n; // 油井数 + float mid; // y数组的中间位置的数 + float minDistance = 0; // 各油井到主管道之间的管道长度总和最小位置 + + cout << "---------------- 输油管问题 ----------------" << endl; + cout << "请输入油井数 n (n>=0) " << endl; + + // 输入油井数 + cin >> n; + + // 判断输入数据合法性，包括检查输入是否为数字，k值是否大于0 + if (cin.fail() || n < 0) + { + cout << "输入n错误！" << endl; + system("pause"); + return 0; + } + + // 分配n个y坐标存储空间 + y = new float[n]; + + cout << "请输入 " << n << " 个油井的 y 坐标（用空格隔开）" << endl; + + // 输入油井的 y 坐标 + for (auto i = 0; i < n; i++) + { + cin >> y[i]; + } + + // 判断输入数据合法性 + if (cin.fail()) + { + cout << "输入y坐标错误！" << endl; + system("pause"); + return 0; + } + + // 运用快速排序对y坐标数组进行排序 + quick_sort(0, n - 1); + + // 计算y数组的中间位置的数 + mid = y[n / 2]; + + // 计算各个油井到主输油管的长度之和 + for (auto i = 0; i < n; i++) + { + minDistance += abs(y[i] - mid); + } + + // 显示输出 + cout << "------------------------------------------------" << endl; + // 判断油井奇偶，做不同的输出 + if (n & 1) + { + // n为奇数，则最优位置为y数组的第n/2个油井的y坐标 + cout << "主管道的最优位置为：y = " << mid << endl; + } + else + { + // n为偶数，则最优位置为y数组的中间两个油井的y坐标的区间 + cout << "主管道的最优位置为：y = [" << y[n / 2 - 1] << "," << mid << "]" << endl; + } + // 输出各油井到主管道之间的管道总长度 + cout << "各油井到主管道之间的管道总长度为：" << minDistance << endl; + cout << "------------------------------------------------" << endl; + + // 暂停查看结果 + system("pause"); + + // 释放内存 + delete[] y; + + // 指针置空 + y = NULL; + + return 0; +} diff --git a/Problems/TubingProblem/Tubing.exe b/Problems/TubingProblem/Tubing.exe new file mode 100644 index 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[蛮力字符串匹配](Algorithm/BruteForceStringMatch.h) +* [文件查找](Algorithm/FileSearch) ## Problems +### Single Problem + +* [Chessboard Coverage Problem (棋盘覆盖问题)](Problems/ChessboardCoverageProblem) + +* [Knapsack Problem (背包问题)](Problems/KnapsackProblem) + +* [Neumann Neighbor Problem (冯诺依曼邻居问题)](Problems/NeumannNeighborProblem) + +* [Round Robin Problem (循环赛日程安排问题)](Problems/RoundRobinProblem) + +* [Tubing Problem (输油管道问题)](Problems/TubingProblem) + ### Leetcode Problems #### Array