经典矩阵快速幂之二-----hdu2157(走k步到

　　题意：(中问题，题意很简单

　　思路：a走k步到b，其实就是A^k，ans.mat[a][b]就是答案。

　　其实就是离散的邻接矩阵那个P(不想证明，逃

#include<cstdio>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<vector>
#include<stack>
#include<cstring>
#include<queue>
#include<set>
#include<string>
#include<map>
#include <time.h>
#define PI acos(-1)
using namespace std;
typedef long long ll;
typedef double db;
const int maxn = 20;
const int N = 10;
const ll maxm = 1e7;
const int INF = 0x3f3f3f;
const int mod=1000;
const ll inf = 1e15 + 5;
const db eps = 1e-9;
int n, m;

struct Matrix{
    int mat[maxn][maxn];
    Matrix operator*(const Matrix& m)const{
        Matrix tmp;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                tmp.mat[i][j]=0;
                for (int k = 0; k < n; k++) {
                    tmp.mat[i][j] += mat[i][k]*m.mat[k][j]%mod;
                    tmp.mat[i][j]+=mod;
                    tmp.mat[i][j] %= mod;
                }
            }
        }
        return tmp;
    }
};

Matrix Pow(Matrix m, int k) {
    Matrix ans;
    memset(ans.mat , 0 , sizeof(ans.mat));
    for (int i=0; i<n; i++) {
        ans.mat[i][i]=1;
    }
    while(k){
        if(k&1)
           ans = ans*m;
        k >>= 1;
        m = m*m;
    }
    return ans;
}

void solve() {
    Matrix tmp;
    while(scanf("%d%d", &n, &m)!=EOF) {
        if (!n&&!m)  break;
        memset(tmp.mat, 0, sizeof(tmp.mat));
        for (int i=0; i<m; i++) {
            int u, v;  scanf("%d%d", &u, &v);
            tmp.mat[u][v]=1;
        }
        int t;  scanf("%d", &t);
        while(t--) {
            Matrix hh;
            int a, b, k;  scanf("%d%d%d", &a, &b, &k);
            hh=Pow(tmp, k);
            printf("%d
", hh.mat[a][b]);
        }
    }

}
int main() {
    int t = 1;
    //freopen("in.txt", "r", stdin);
   // scanf("%d", &t);
    while(t--)
        solve();
    return 0;
}