POJ 1681 Painter's Problem 高斯消元

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
", 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;
}