POJ 1830 开关问题 高斯消元

基础的高斯消元解决异或方程问题。

#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 = 45;

int a[maxn][maxn], n, s1[maxn], e1[maxn];

void Gauss() {
    for(int i = 0; i < n; i++) {
        int k = i;
        while(a[k][i] == 0 && k < n) k++;
        if(k >= n) continue;
        for(int j = 0; j <= n; j++) swap(a[i][j], a[k][j]);
        for(int j = 0; j < n; j++) if(i != j && a[j][i] != 0) {
            for(int k = 0; k <= n; k++) {
                a[j][k] ^= a[i][k];
            }
        }
    }
}

void pa() {
    for(int i = 0; i < n; i++) {
        for(int j = 0; j <= n; j++) {
            printf("%d ", a[i][j]);
        }
        puts("");
    }
}

int main() {
    int T; scanf("%d", &T);
    while(T--) {
        memset(a, 0, sizeof(a));
        scanf("%d", &n);
        for(int i = 0; i < n; i++) scanf("%d", &s1[i]);
        for(int i = 0; i < n; i++) scanf("%d", &e1[i]);
        for(int i = 0; i < n; i++) if(s1[i] != e1[i]) a[i][n] = 1;
        for(int i = 0; i < n; i++) a[i][i] = 1;
        int ia, ib;
        while(scanf("%d%d", &ia, &ib)) {
            if(ia == 0 && ib == 0) break;
            ia--; ib--;
            a[ib][ia] = 1;
        }
        Gauss();
        bool bad = false;
        int fcnt = 0, ans = 0;
        for(int i = 0; i < n; i++) {
            int nowsum = 0;
            for(int j = 0; j < n; j++) nowsum += a[i][j];
            if(nowsum == 0 && a[i][n] != 0) bad = true;
            if(a[i][n] == 0 && nowsum == 0) fcnt++;
        }
        //printf("the answer is :");
        if(bad) puts("Oh,it's impossible~!!");
        else cout << (1LL << fcnt) << endl;
    }
    return 0;
}