AcWing - 208 - 开关问题 = 高斯消元法

Posted 2022-11-05 inko

https://www.acwing.com/problem/content/210/

要注意两点：开关之间的关系不一定是对称的，并且每个开关会控制自己。

消元的过程中可以计算出矩阵的秩，假如某个行没有主元但是有常数，那么就直接-1了。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

const int MAXN = 40;

bool a[MAXN + 5][MAXN + 1 + 5 ];
bool ans[MAXN + 5];

int Gauss_Jordan(int n) 
    int r = 0;
    for(int i = 1; i <= n; ++i) 
        int mx = i;
        for(int j = i; j <= n; ++j) 
            if(a[j][i]) 
                mx = j;
                break;
            
        

        if(!a[mx][i])
            continue;

        ++r;
        if(mx != i) 
            for(int j = 1; j <= n + 1; ++j)
                swap(a[i][j], a[mx][j]);
        

        for(int j = 1; j <= n + 1; ++j) 
            if(j != i && a[j][i]) 
                for(int k = 1; k <= n + 1; ++k)
                    a[j][k] ^= a[i][k];
            
        
    
    for(int i = 1; i <= n; ++i) 
        if(a[i][i] == 0 && a[i][n + 1])
            return -1;
    

    /*for(int i = 1; i <= n; ++i) 
        for(int j = 1; j <= n + 1; ++j) 
            printf(" %d", (int)a[i][j]);
        
        printf("\n");
    
    puts("---\n");*/

    for(int i = 1; i <= n; ++i)
        ans[i] = a[i][n + 1] & a[i][i];
    return r;


int main() 
#ifdef Yinku
    freopen("Yinku.in", "r", stdin);
#endif // Yinku
    int T;
    scanf("%d", &T);
    while(T--) 
        memset(a, 0, sizeof(a));
        int n;
        scanf("%d", &n);
        for(int i = 1, x; i <= n; ++i) 
            scanf("%d", &x);
            a[i][n + 1] = x;
        
        for(int i = 1, x; i <= n; ++i) 
            scanf("%d", &x);
            a[i][n + 1] ^= x;
        
        for(int i = 1; i <= n; ++i)
            a[i][i] = 1;

        int i, j;
        do 
            scanf("%d%d", &i, &j);
            if(i == 0 && j == 0)
                break;
            a[j][i] = 1;
         while(1);

        /*for(int i = 1; i <= n; ++i) 
            for(int j = 1; j <= n + 1; ++j) 
                printf(" %d", (int)a[i][j]);
            
            printf("\n");
        
        puts("---\n");*/

        int r = Gauss_Jordan(n);
        if(r == -1)
            puts("Oh,it's impossible~!!");
        else
            printf("%lld\n", 1ll << (n - r));