ZOJ 3497--Mistwald

设有向图$D=<V,E>,V=\{v_1,v_2,\dots ,v_n\}$，令$a_{ij}^{(1)}$顶点$v_i$邻接到顶点$v_j$的边的条数，称$(a_{ij}^{(1)}{)}_{n\times n}$为$D$的邻接矩阵，简记为$A$。
定理：$A$的$l$次幂$A^l(l\ge 1)$中元素$a_{ij}^{(l)}$为$D$中$v_i$到$v_j$长度为$l$的通路数。

对于此题，采用类邻接矩阵，只需记录$v_i$是否可以到达$v_j$，即边数为$1$。
对于每次询问，跑矩阵快速幂判断即可。

AC代码：

#include 
using namespace std;

typedef long long ll;

struct mat
{
    ll a[50][50];
};

int szie;
mat A, B, V;

mat mat_mul(mat x, mat y)
{
    mat res;
    memset(res.a, 0, sizeof(res.a));
    for(int i = 1; i <= szie; i++){
        for(int j = 1; j <= szie; j++){
            for(int k = 1; k <= szie; k++){
                res.a[i][j] = (res.a[i][j] + x.a[i][k] * y.a[k][j]);
            }
        }
    }
    return res;
}

mat pow_matrix_mod(int n)
{
    mat c = A, ans = V;
    while(n){
        if(n & 1)   ans = mat_mul(ans, c);
        c = mat_mul(c, c);
        n >>= 1;
    }
    return ans;
}

int main()
{
    int T, n, m;
    cin >> T;
    while (T--) {
        memset(V.a, 0, sizeof(V.a));
        memset(A.a, 0, sizeof(A.a));
        memset(B.a, 0, sizeof(B.a));
        scanf("%d %d", &n, &m);
        getchar();
        szie = n*m;
        for (int i = 1; i <= szie; ++i) V.a[i][i] = 1;
        int x1, y1, x2, y2, x3, y3, x4, y4;
        for (int i = 1; i <= n; ++i) {
            for (int j = 1; j <= m; ++j) {
                scanf("((%d,%d),(%d,%d),(%d,%d),(%d,%d))",&x1, &y1, &x2, &y2, &x3, &y3, &x4, &y4);
                if (i == n && j == m)   continue;
                int p = (i-1)*m+j;
                A.a[p][(x1-1)*m+y1] = 1;
                A.a[p][(x2-1)*m+y2] = 1;
                A.a[p][(x3-1)*m+y3] = 1;
                A.a[p][(x4-1)*m+y4] = 1;
                getchar();
            }
        }
        int Q, l;
        scanf("%d", &Q);
        while (Q--) {
            scanf("%d", &l);
            B = pow_matrix_mod(l);
            if (B.a[1][szie] == 0) cout << "False" << '\n';
            else {
                int flag = 0;
                for (int i = 1; i < szie; ++i) {
                    if (B.a[1][i]) {
                        flag = 1;
                        break;
                    }
                }
                if (!flag)  cout << "True" << '\n';
                else    cout << "Maybe" << '\n';
            }
        }
        cout << '\n' ;
    }
    return 0;
}