How many ways??

http://acm.hdu.edu.cn/showproblem.php?pid=2157

题解：经典矩阵算法。把给定的图转为邻接矩阵，即A(i,j)=1当且仅当存在一条边i->j。令C=A*A，那么C(i,j)=ΣA(i,k)*A(k,j)，实际上就等于从点i到点j恰好经过2条边的路径数（枚举k为中转点）。类似地，C*A的第i行第j列就表示从i到j经过3条边的路径数。同理，假设要求经过k步的路径数，我们仅仅须要二分求出A^k就可以。

参考文章：矩阵快速幂

/*
*@Author:   STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=100000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,p,l,r,u,v;
int ans,cnt,flag,temp,sum;
int s[25][25];
int b[25][25];
int a[25][25];
char str;
struct node{};
void Matrix(int a[25][25],int b[25][25]){int c[25][25];memset(c,0,sizeof(c));for(int i=0;i<n;i++){for(int j=0;j<n;j++){for(int k=0;k<n;k++){c[i][j]=(c[i][j]+a[i][k]*b[k][j])%1000;}}}for(int i=0;i<n;i++){for(int j=0;j<n;j++){a[i][j]=c[i][j];}}
}
int power(int A,int B,int k){for(int i=0;i<n;i++){for(int j=0;j<n;j++){a[i][j]=(i==j);b[i][j]=s[i][j];}}while(k){if(k&1)Matrix(a,b);Matrix(b,b);k>>=1;}return a[A][B];
}
int main()
{
#ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout);
#endif//ios::sync_with_stdio(false);//cin.tie(0);//cout.tie(0);//while(~scanf("%d%d",&n,&m)&&n+m){memset(s,0,sizeof(s));for(int i=1;i<=m;i++){scanf("%d%d",&u,&v);s[u][v]=1;}scanf("%d",&t);for(int i=1;i<=t;i++){scanf("%d%d%d",&u,&v,&k);cout<<power(u,v,k)<<endl;}}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif//cout << "Hello world!" << endl;return 0;
}

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