1. 题目

1.1 题目链接

http://poj.org/problem?id=1753

1.2 题目大意

一个黑白棋子的棋盘，一个反过来周围四个也跟着反过来(如果存在的话)，颜色取反，问最少反转次数使得颜色全白或者全黑，不存在解的话输出信息。

1.3 解题思路

采用回溯算法，暴力枚举

2. 代码

2.1 Wrong Answer代码

/*** @description: * @author: michael ming* @date: 2019/7/11 22:09* @modified by: */
#include <iostream>
#include <string>
using namespace std;
bool a[4][4];
bool isok()//判断是否都是同种颜色
{int i, j;for(i = 0; i < 4; ++i){for(j = 0; j < 4; ++j){if(a[i][j] != a[0][0]){return false;}}}return true;
}
void flipAndUpdate(int r, int c)//翻转r，c处及其周围棋子
{a[r][c] = !a[r][c];if(r-1 >= 0)a[r-1][c] = !a[r-1][c];if(r+1 < 4)a[r+1][c] = !a[r+1][c];if(c-1 >= 0)a[r][c-1] = !a[r][c-1];if(c+1 < 4)a[r][c+1] = !a[r][c+1];
}
void flip(int r, int c,int curstep, long &minstep)
{if(isok()){if(curstep < minstep)minstep = curstep;return;}if(c+1 < 4)flip(r,c+1,curstep,minstep);else if(c+1 == 4 && r+1 < 4)flip(r+1,0,curstep,minstep);flipAndUpdate(r,c);curstep++;if(c+1 < 4)flip(r,c+1,curstep,minstep);else if(c+1 == 4 && r+1 < 4)flip(r+1,0,curstep,minstep);flipAndUpdate(r,c);//翻完了，还要复原？
}
int main()
{string s;int i, j;long minstep = 65536;for(i = 0; i < 4; ++i){cin >> s;for(j = 0; j < 4; ++j){if(s[j] == 'b')a[i][j] = 1;elsea[i][j] = 0;}}flip(0,0,0,minstep);if(minstep == 65536)cout << "Impossible" << endl;elsecout << minstep;return 0;
}

2.2 Accepted代码

上面代码范围有问题，改动如下

AC 代码如下

/*** @description: * @author: michael ming* @date: 2019/7/11 22:09* @modified by: */
#include <iostream>
#include <string>
using namespace std;
bool a[5][5];
bool isok()//判断是否都是同种颜色
{int i, j;for(i = 0; i < 4; ++i){for(j = 0; j < 4; ++j){if(a[i][j] != a[0][0]){return false;}}}return true;
}
void flipAndUpdate(int r, int c)//翻转r，c处及其周围棋子
{a[r][c] = !a[r][c];if(r-1 >= 0)a[r-1][c] = !a[r-1][c];if(r+1 < 4)a[r+1][c] = !a[r+1][c];if(c-1 >= 0)a[r][c-1] = !a[r][c-1];if(c+1 < 4)a[r][c+1] = !a[r][c+1];
}void flip(int r, int c,int curstep, long &minstep)
{if(isok()){if(curstep < minstep)minstep = curstep;return;}if(r == 4)return;if(c+1 < 4)flip(r,c+1,curstep,minstep);else if(c+1 == 4)flip(r+1,0,curstep,minstep);flipAndUpdate(r,c);curstep++;if(c+1 < 4)flip(r,c+1,curstep,minstep);else if(c+1 == 4)flip(r+1,0,curstep,minstep);flipAndUpdate(r,c);//翻完了，还要复原？
}
int main()
{string s;int i, j;long minstep = 65536;for(i = 0; i < 4; ++i){cin >> s;for(j = 0; j < 4; ++j){if(s[j] == 'b')a[i][j] = 1;elsea[i][j] = 0;}}flip(0,0,0,minstep);if(minstep == 65536)cout << "Impossible" << endl;elsecout << minstep;return 0;
}