zoj3497Mistwald矩阵

　　判断i到j    是否k步可达。

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <string.h>
typedef long long LL;
using namespace std;
int n;
int x[100], y[100];
struct Matrix
{
    int m[40][40];
};

Matrix Mul(Matrix a, Matrix b)
{
    Matrix ans;
    for (int i = 0; i < n; i++){
        for (int j = 0; j < n; j++){
            ans.m[i][j] = 0;
            for (int k = 0; k < n; k++){
                ans.m[i][j] |= a.m[i][k] * b.m[k][j];
            }
        }
    }
    return ans;
}

Matrix quick(Matrix a, int b)
{
    Matrix  ans;
    for (int i = 0; i < n;i++)
    for (int j = 0; j < n; j++)
        ans.m[i][j] = (i == j);
    while (b){
        if (b & 1) ans = Mul(ans, a);
        a = Mul(a, a);
        b >>= 1;
    }
    return ans;
}

Matrix init()
{
    int a, b;
    Matrix ans;
    cin >> a >> b;
    n = a*b;
    for (int i = 0; i < n;i++)
    for (int j = 0; j < n; j++)
        ans.m[i][j] = 0;
    for (int i = 0; i < a; i++){
        for (int j = 0; j < b; j++){
            getchar();
            scanf("((%d,%d),(%d,%d),(%d,%d),(%d,%d))", &x[0], &y[0], &x[1], &y[1], &x[2], &y[2], &x[3], &y[3]);
            if(i==a-1&&j==b-1)continue;//图中路径不能经过终点
            int t = i*b + j;
            for (int k = 0; k < 4; k++){
                x[k]--; y[k]--;
                int t1 = x[k] * b + y[k];
                ans.m[t][t1] = 1;
            }
        }
    }
    return ans;
}

void gao(Matrix a, int b)
{
    int flag = 0;
    int flag1 = 0;
    Matrix ans;
    ans = quick(a, b);
    if(!ans.m[0][n-1]){
        printf("False
"); return ;
    }
    for(int i = 1;i<n-1;i++)
    if(ans.m[0][i]){
        printf("Maybe
");return ;
    }
    printf("True
");
}
int main()
{
    int T; int q; int t;
    cin >> T;
    while (T--){
        Matrix ans = init();
        cin >> q;
        while(q--){
            scanf("%d", &t);
            gao(ans, t);
        }
        cout<<endl;
    }
    return 0;
}