标签:blog io os ar for sp div on log
利用高斯消元求解异或方程
#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; 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(""); } } void solve() { 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(); int ans = 0; bool bad = false; for(int i = 0; i < n * n; i++) { if(a[i][n * n]) ans++; 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(bad) puts("inf"); else printf("%d\n", ans); } int main() { int T; scanf("%d", &T); while(T--) { memset(a, 0, sizeof(a)); scanf("%d", &n); for(int i = 0; i < n; i++) { for(int j = 0; j < n; j++) { char tmp; scanf(" %c", &tmp); if(tmp == ‘w‘) a[i * n + j][n * n] = 1; else a[i * n + j][n * n] = 0; } } solve(); } return 0; }
POJ 1681 Painter's Problem 高斯消元
标签:blog io os ar for sp div on log
原文地址:http://www.cnblogs.com/rolight/p/4066464.html