uva 11464 偶数矩阵

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题意：给定一个矩阵，其中元素为 0或者1

花最少次数改变其中的0 到1，使得每个元素上下左右的数字之和为偶数

方法：枚举第一行2^n 种情况，最后判断最后一行是否满足

#include <iostream>#include <vector>#include <map>#include <list>#include <set>#include <deque>#include <stack>#include <queue>#include <algorithm>#include <cmath>#include <cctype>#include <cstdio>#include <iomanip>#include <cmath>#include <cstdio>#include <iostream>#include <string>#include <sstream>#include <cstring>#include <queue>using namespace std;///宏定义const int INF = 20000000;const int maxn = 20;///全局变量 和 函数//int T;int board[maxn][maxn];int board1[maxn][maxn];int n;int cnt;bool test(){int i, j;for (i = 1; i <= n - 1; i++){for (j = 1; j <= n; j++){//推导下一行的状态if ((board[i - 1][j] + board[i + 1][j] + board[i][j - 1] + board[i][j + 1]) % 2 != 0){if (board[i + 1][j] == 1)return false;else{board[i + 1][j] = 1;cnt++;}}}}//最后一行for (i = 1; i <= n; i++){if ((board[n - 1][i] + board[n + 1][i] + board[n][i - 1] + board[n][i + 1]) % 2 != 0)return false;}return true;}int main(){//////////////////////////////////////////////////////////////////////////int i, j;scanf("%d", &T);int cases = 1;while (T--){memset(board, 0, sizeof(board));scanf("%d", &n);for (i = 1; i <= n; i++){for (j = 1; j <= n; j++){scanf("%d", &board[i][j]);board1[i][j] = board[i][j];}}int ans = INF;for (i = 0; i < (1 << n); i++){//复原for (int nn = 1; nn <= n; nn++){for (int mm = 1; mm <= n; mm++){board[nn][mm] = board1[nn][mm];}}cnt = 0;//第一行的状态枚举for (j = 0; j < n; j++){if ((i & (1 << j)) && (board[1][n - j] == 0) ){board[1][n - j] = 1;cnt++;}}//测试if (test()){ans = min(cnt, ans);}}if (ans == INF){printf("Case %d: %d\n", cases++, -1);}elseprintf("Case %d: %d\n", cases++, ans);}return 0;}