题目大意：给出灯的一些关系，求有多少种方法从始状态到终状态。
思路：其实根据灯的这些关系就可以列出一系列方程，，然后用高斯消元就可以求出自由元的数量，答案就是2^自由元的数量。
CODE：#define _CRT_SECURE_NO_WARNINGS#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#define MAX 50using namespace std;int cnt;int a[MAX][MAX];inline int Gauss(){int re = 0,now = 1;for(int j,i = 1; i <= cnt; ++i) {for(j = now; j <= cnt && !a[j][i]; ++j);if(j > cnt) {++re;continue;}for(int k = 1; k <= cnt + 1; ++k)swap(a[now][k],a[j][k]);for(j = now + 1; j <= cnt; ++j)if(a[j][i])for(int k = i; k <= cnt + 1; ++k)a[j][k] ^= a[now][k];++now;}for(int i = now; i <= cnt; ++i)if(a[i][cnt + 1])return -1;return 1 << re;}int main(){int T;for(cin >> T; T–;) {memset(a,0,sizeof(a));scanf("%d",&cnt);for(int i = 1; i <= cnt; ++i)scanf("%d",&a[i][cnt + 1]);for(int x,i = 1; i <= cnt; ++i) {scanf("%d",&x);a[i][cnt + 1] ^= x;}int x,y;while(scanf("%d%d",&x,&y),x + y)a[y][x] = 1;for(int i = 1; i <= cnt; ++i)a[i][i] = 1;int ans = Gauss();if(ans == -1)puts("Oh,it's impossible~!!");elseprintf("%d\n",ans);}return 0;}
谁是谁生命的点缀。
