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#include <iostream>
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#include <string.h>
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#include <stdio.h>
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#include <vector>
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#include <map>
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#include <queue>
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#include <algorithm>
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#include <math.h>
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#include <cstdio>
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using namespace std;
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const int N = 30;
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const double eps = 10e-8;
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int n;
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int start[N]; //开始状态
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int stop[N]; //结束状态
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//高斯消元模板
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double a[N][N]; //增广矩阵
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int gauss() {
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int c, r; //当前列,行
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// 1、枚举系数每一列
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for (c = 0, r = 0; c < n; c++) {
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// 2、找出系数最大的行
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int t = r;
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for (int i = r; i < n; i++) //行
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if (abs(a[i][c]) > abs(a[t][c])) t = i;
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// 3、最大系数为0，直接下一列
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if (abs(a[t][c]) < eps) continue;
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// 4、交换
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if (r != t) // POJ中，如果二维数组，直接swap(a[t],a[r])会报编译错误，没办法，只好用了循环
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for (int i = 0; i < N; i++) swap(a[t][i], a[r][i]);
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// 5、倒序, 每项除a[r][c],化系数为1,处理的是方程左右两端，需要带着a[r][n]
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for (int i = n; i >= c; i--) a[r][i] /= a[r][c];
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// 6、用当前行将下面所有的列消成0
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for (int i = r + 1; i < n; i++)
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for (int j = n; j >= c; j--)
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a[i][j] -= a[r][j] * a[i][c];
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// 7、下一行
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r++;
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}
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if (r < n) {
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for (int i = r; i < n; i++)
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if (abs(a[i][n]) > eps)
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return -1; //无解
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return n - r; //自由元个数
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}
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//倒三角，将已知解代入
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for (int i = n - 2; i >= 0; i--)
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for (int j = i + 1; j < n; j++)
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a[i][n] -= a[i][j] * a[j][n];
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//唯一解：0
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return 0;
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}
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int main() {
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int T;
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cin >> T;
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while (T--) {
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cin >> n;
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memset(a, 0, sizeof(a));
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memset(start, 0, sizeof(start)); //初始化开始状态
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memset(stop, 0, sizeof(stop)); //初始化终止状态
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//输入起始状态(下标从0开始)
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for (int i = 0; i < n; i++) cin >> start[i];
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//输入终止状态(下标从0开始)
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for (int i = 0; i < n; i++) cin >> stop[i];
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//输入增广矩阵
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int x, y;
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while (cin >> x >> y && x != 0 && y != 0)
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a[y - 1][x - 1] = 1; //反着存入y-1受x-1影响
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for (int i = 0; i < n; i++) {
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a[i][n] = start[i] ^ stop[i]; //状态变化 start^stop
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a[i][i] = 1; //自己影响自己
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}
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//高斯消元模板
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int t = gauss();
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if (t == -1)
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cout << "Oh,it's impossible~!!" << endl;
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else
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cout << (1 << t) << endl;
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}
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return 0;
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} |
