搜索 - poj1753 Flip Game

题目链接：

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

题意：

4*4棋盘，每个格子上黑或白棋子一枚，翻转棋子时相邻的四个棋子都会随之改变颜色，问最小翻转步数，若无输出“Impossible”

思路：

对于这种翻转游戏，翻转顺序不会影响结果，直接枚举即可，时间复杂度2^16

核心算法是从16个数中分别选出1.2.3.....16个数

代码：

#include <stdio.h>int check = 0;int isFinish(int arr[6][6], int sum){int c = 1;for (int i = 1; i <= 4&&c; i++){for (int j = 1; j <= 4&&c; j++){if (arr[i][j] == 1){c = 0;}}}if (c){check = sum;return 1;}else{c = 1;for (int i = 1; i <= 4 && c; i++){for (int j = 1; j <= 4 && c; j++){if (arr[i][j] == 0){c = 0;}}}if (c){check = sum;return 1;}}return 0;}void search(int arr[6][6], int pos,int x,int sum){if (check) return;int i = (pos - 1) / 4 + 1, j = pos % 4;j = j == 0 ? 4 : j;arr[i][j] = 1 - arr[i][j];arr[i + 1][j] = 1 - arr[i + 1][j];arr[i][j + 1] = 1 - arr[i][j + 1];arr[i - 1][j] = 1 - arr[i - 1][j];arr[i][j - 1] = 1 - arr[i][j - 1];if (x != 1){for (int k = pos + 1; k <= 16 - (x - 1) + 1; k++)search(arr, k, x - 1, sum);}else{isFinish(arr, sum);}arr[i][j] = 1 - arr[i][j];arr[i + 1][j] = 1 - arr[i + 1][j];arr[i][j + 1] = 1 - arr[i][j + 1];arr[i - 1][j] = 1 - arr[i - 1][j];arr[i][j - 1] = 1 - arr[i][j - 1];return;}int main(){int arr[6][6] = { 0 };char str[5];for (int i = 1; i <= 4; ++i){scanf("%s", str);int t = 0;for (int j = 0; j < 4; ++j){if (str[j] == 'w')//b=0,w=1;arr[i][j+1] = 1;}}//inif (isFinish(arr, 0)) printf("0\n");else{for (int i = 1; i <= 16 && !check; ++i){for (int j = 1; j <= 16 - i + 1 && !check; ++j){search(arr, j, i, i);}}if (check) printf("%d\n", check);else printf("Impossible\n");}return 0;}