先用高斯消元化简,然后枚举自由变元
#include <cstdio> #include <cstring> #include <algorithm> #include <queue> #include <stack> #include <map> #include <set> #include <climits> #include <iostream> #include <string> using namespace std; #define MP make_pair #define PB push_back typedef long long LL; typedef unsigned long long ULL; typedef vector<int> VI; typedef pair<int, int> PII; typedef pair<double, double> PDD; const int INF = INT_MAX / 3; const double eps = 1e-8; const LL LINF = 1e17; const double DINF = 1e60; const int maxn = 250; const int dx[] = {0, 0, 1, -1}; const int dy[] = {1, -1, 0, 0}; int a[maxn][maxn], n; bool fx[maxn]; char buf[16][16]; void Gauss() { for(int i = 0; i < n * n; i++) { int k = i; while(a[k][i] == 0 && k < n * n) k++; if(k >= n * n) break; for(int j = 0; j <= n * n; j++) swap(a[i][j], a[k][j]); for(int j = 0; j < n * n; j++) if(j != i && a[j][i] != 0) { for(int k = 0; k <= n * n; k++) { a[j][k] ^= a[i][k]; } } } } void pp() { for(int i = 0; i < n * n; i++) { for(int j = 0; j <= n * n; j++) { printf("%d ", a[i][j]); } puts(""); } } int solve() { memset(a, 0, sizeof(a)); for(int i = 0; i < n; i++) { for(int j = 0; j < n; j++) { if(buf[i][j] == 'w') a[i * n + j][n * n] = 1; else a[i * n + j][n * n] = 0; } } for(int i = 0; i < n; i++) { for(int j = 0; j < n; j++) { int u = i * n + j; a[u][u] = 1; for(int k = 0; k < 4; k++) { int nx = i + dx[k], ny = j + dy[k], nu = nx * n + ny; if(nx >= 0 && nx < n && ny >= 0 && ny < n) { a[u][nu] = a[nu][u] = 1; } } } } Gauss(); memset(fx, 0, sizeof(fx)); int ans = INF, fcnt = 0; bool bad = false; for(int i = 0; i < n * n; i++) { int nsum = 0; for(int j = 0; j < n * n; j++) nsum += a[i][j]; if(nsum == 0 && a[i][n * n] == 1) bad = true; if(nsum == 0) fx[i] = true, fcnt++; } if(bad) return INF; for(int s = 0; s < (1 << fcnt); s++) { int nowans = 0; for(int i = 0; i < n * n - fcnt; i++) { int col = 0; for(int j = 0; j < n * n; j++) if(fx[j] && a[i][j]) { if(s & (1 << (n * n - j - 1))) col ^= 1; } if(a[i][n * n] != col) nowans++; } int bcnt = 0; for(int i = 0; i < fcnt; i++) if(s & (1 << i)) bcnt++; ans = min(nowans + bcnt, ans); } return ans; } int main() { while(scanf("%s%s%s%s", buf[0], buf[1], buf[2], buf[3]) != EOF) { n = 4; int ans = solve(); for(int i = 0; i < 4; i++) { for(int j = 0; j < 4; j++) { if(buf[i][j] == 'w') buf[i][j] = 'b'; else buf[i][j] = 'w'; } } ans = min(ans, solve()); if(ans < INF) printf("%d ", ans); else puts("Impossible"); } return 0; }