HDU-2157 How many ways??

题目说明可以走重复边，用矩阵快速幂即可求出，i->j的方案 = i->k的方案*k->j的方案。

注意输入数据存在多条相同的边。

#include 
#include 
using namespace std;
const int N = 1e2;
const int mod = 1e3;
struct abc{
	int t[N][N];
}a, ans, tmp;

void init1(int n){
	memset(ans.t, 0, sizeof(ans.t));
	memset(tmp.t, 0, sizeof(tmp.t));
	memset(a.t, 0, sizeof(a.t));
	for(int i = 0; i < n; ++i){
		ans.t[i][i] = 1;
	}
}

void init2(int n){
	memset(ans.t, 0, sizeof(ans.t));
	for(int i = 0; i < n; ++i){
		ans.t[i][i] = 1;
	}
	for(int i = 0; i < n; ++i){
		for(int j = 0; j < n; ++j){
			a.t[i][j] = tmp.t[i][j];
		}
	}
}

struct abc x(struct abc a, struct abc b, int n){
	struct abc c;
	memset(c.t, 0, sizeof(c.t));
	for(int i = 0; i < n; ++i){
		for(int j = 0; j < n; ++j){
			for(int k = 0; k < n; ++k){
				c.t[i][j] = (c.t[i][j] +  (a.t[i][k] * b.t[k][j]) % mod) % mod;
			}
		}
	}
	return c;
}
int main()
{
	int n, m;
	while(scanf("%d %d", &n, &m) && n + m != 0){
		init1(n);
		for(int i = 0; i < m; ++i){
			int u, v;
			scanf("%d %d", &u, &v);
			a.t[u][v] = 1;
			tmp.t[u][v] = 1;
		}
		int h;
		scanf("%d", &h);
		for(int i = 0; i < h; ++i){
			int u, v, d;
			scanf("%d %d %d", &u, &v, &d);
			while(d){
				if(d & 1){
					ans = x(ans, a, n);
				}
				a = x(a, a, n);
				d = d >> 1;
			}
			printf("%d\n", ans.t[u][v]);
			init2(n);
		}
	}
	return 0;
}```