POJ 1830.开关问题（高斯消元）

题目链接

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Solutin：

           将每个开关使用的情况当成未知数，如果开关i能影响到开关j，那么系数矩阵A[j][i]的系数为1。

           每个开关增广矩阵的值是开关k的初状态异或开关k的目标状态，这个应该很容易想到。

           方程都列好了，直接消元就好了。

code

/*
       解异或方程组
*/
#include <iostream>
#include <cstring>
using namespace std;
const int MAXN = 50;

int prim[MAXN];
int A[MAXN][MAXN];

int Gauss (int n, int m) {
    int col = 0, row = 0, tem, k = 0;
    for (; col < m && row < n; ++col) {
        for (tem = row; tem < n && A[tem][col] == 0; ++tem);
        if (tem == n)  continue;
        if (tem != row) {
            for (int i = 0; i <= n; ++i)
                swap (A[row][i], A[tem][i]);
        }
        for (int i = row + 1; i <= n; ++i) {
            if (A[i][col])
                for (int j = col; j <= m; ++j)
                    A[i][j] ^= A[row][j];
        }
        row++;
    }
    for (int i = row; i < n; ++i)
        if (A[i][n] != 0) return -1;
    return m - row;
}
int n, m, cs, S, T;
int main() {
    ios::sync_with_stdio (0);
    cin >> cs;
    while (cs--) {
        cin >> n;
        S = T = 0;
        for (int i = 0, x; i < n; ++i) {
            cin >> x;
            if (x) S |= (1 << i);
        }
        for (int i = 0, x; i < n; ++i) {
            cin >> x;
            if (x) T |= (1 << i);
        }
        T ^= S;
        memset (A, 0, sizeof A);
        for (int i = 0; i < n; ++i){
            if (T & (1 << i) ) A[i][n] = 1;
            A[i][i]=1;
        }
        int x, y;
        while (cin >> x >> y, x && y)   A[y - 1][x - 1] = 1;
        int p = Gauss (n, n);
        if (p < 0) cout << "Oh,it's impossible~!!" << endl;
        else
            cout << (1 << p) << endl;
    }
}