HDU 2157 How many ways?? 矩阵

可达矩阵的K次幂便是从i到j走K步能到达的方案数。

注意处理k=0的情况。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <climits>
#include <iostream>
#include <string>

using namespace std;
 
#define MP make_pair
#define PB push_back
typedef long long LL;
typedef unsigned long long ULL;
typedef vector<int> VI;
typedef pair<int, int> PII;
typedef pair<double, double> PDD;
const int INF = INT_MAX / 3;
const double eps = 1e-8;
const LL LINF = 1e17;
const double DINF = 1e60;
const int maxn = 30;
const int mod = 1000;

struct Matrix {
    int n, m, data[maxn][maxn];
    Matrix(int n = 0, int m = 0): n(n), m(m) {
        memset(data, 0, sizeof(data));
    }
};

Matrix operator * (Matrix a, Matrix b) {
    Matrix ret(a.n, b.m);
    for(int i = 1; i <= a.n; i++) {
        for(int j = 1; j <= b.m; j++) {
            for(int k = 1; k <= a.m; k++) {
                ret.data[i][j] += a.data[i][k] * b.data[k][j];
                ret.data[i][j] %= mod;
            }
        }
    }
    return ret;
}

Matrix pow(Matrix mat, int p) {
    if(p == 0) {
        Matrix ret(mat.n, mat.m);
        for(int i = 1; i <= mat.n; i++)  ret.data[i][i] = 1;
        return ret;
    }
    if(p == 1) return mat;
    Matrix ret = pow(mat * mat, p / 2);
    if(p & 1) ret = ret * mat;
    return ret;
}

int n, m;

int main() {
    while(scanf("%d%d", &n, &m), n) {
        Matrix mat(n, n);
        for(int i = 1; i <= m; i++) {
            int a, b; scanf("%d%d", &a, &b);
            mat.data[a + 1][b + 1] = 1;
        }
        int Q; scanf("%d", &Q);
        while(Q--) {
            int a, b, k; scanf("%d%d%d", &a, &b, &k);
            Matrix ret = pow(mat, k);
            printf("%d
", ret.data[a + 1][b + 1]);
        }
    }
    return 0;
}