UVa 11464 Even Parity / 深搜

来源：互联网发布：电子音乐专业制作软件编辑：程序博客网时间：2024/06/06 09:57

算是一类的题目 zoj也看到过今天终于写了

给你一个0 1 的矩阵可以把0变成1 1 不能变成0 然后最小的变换次数是每一个位置的上下左右加起来的和是偶数

枚举第一行根据第一行下面的都已经确定了 O(2^n*N*N)

#include <stdio.h>#include <string.h>const int MAX = 20;int n;int min;int a[MAX][MAX];int b[MAX][MAX];int check(){int i,j,sum,cnt = 0;for(i = 0;i < n; i++){if(a[0][i] == 1 && b[0][i] == 0){return 0x7ffffff;}else if(a[0][i] != b[0][i])cnt++;}for(i = 1; i < n; i++){for(j = 0; j < n; j++){sum = 0;if(i > 1)sum += b[i-2][j];if(j > 0)sum += b[i-1][j-1];if(j < n - 1)sum += b[i-1][j+1];b[i][j] = sum % 2;if(a[i][j] == 1 && b[i][j] == 0)return 0x7ffffff;else if(a[i][j] != b[i][j])cnt++;}}return cnt;}void dfs(int k){if(k == n){int res = check();if(min > res)min = res;return;}b[0][k] = 0;dfs(k+1);b[0][k] = 1;dfs(k+1);}int main(){int t,i,j,cas = 1;scanf("%d",&t);while(t--){scanf("%d",&n);for(i = 0; i < n; i++)for(j = 0; j < n; j++)scanf("%d",&a[i][j]);min = 0x7ffffff;dfs(0);printf("Case %d: ",cas++);if(min == 0x7ffffff)puts("-1");elseprintf("%d\n",min);}return 0;}