hdu 2157 How many ways?? (矩阵快速幂经典题目)
题意：给定一个有向图，问从A点恰好走k步（允许重复经过边）到达B点的方案数mod p的值
思路： ** 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，CA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，如果要求经过k步的路径数，我们只需要快速幂求出A^k即可。**
这个转化好美啊。。。
/* ***********************************************
Author :111qqz
Created Time :Wed 19 Oct 2016 08:14:56 PM CST
File Name :code/hdu/2157.cpp
************************************************ */
1#include <cstdio>
2#include <cstring>
3#include <iostream>
4#include <algorithm>
5#include <vector>
6#include <queue>
7#include <set>
8#include <map>
9#include <string>
10#include <cmath>
11#include <cstdlib>
12#include <ctime>
13#define fst first
14#define sec second
15#define lson l,m,rt<<1
16#define rson m+1,r,rt<<1|1
17#define ms(a,x) memset(a,x,sizeof(a))
18typedef long long LL;
19#define pi pair < int ,int >
20#define MP make_pair
1using namespace std;
2const double eps = 1E-8;
3const int dx4[4]={1,0,0,-1};
4const int dy4[4]={0,-1,1,0};
5const int inf = 0x3f3f3f3f;
6const int N=25;
7const int mod = 1000;
8int n,m;
9struct Mat
10{
11 int mat[N][N];
12 void clear()
13 {
14 ms(mat,0);
15 }
16}M;
1Mat operator * (Mat a,Mat b)
2{
3 Mat ans;
4 ans.clear();
5 for ( int i = 0 ; i < n ; i++)
6 for ( int j = 0 ; j < n ;j++)
7 for ( int k = 0 ; k < n ; k++)
8 ans.mat[i][j] = (ans.mat[i][j] + a.mat[i][k] * b.mat[k][j])%mod;
return ans;
}
1Mat operator ^ (Mat a,int b)
2{
3 Mat ans;
4 ans.clear();
5 for ( int i = 0 ; i < n ; i++) ans.mat[i][i] = 1;
1 while (b>0)
2 {
3 if (b&1) ans = ans * a;
4 b = b >> 1LL;
5 a = a* a;
6 }
7 return ans;
8}
9LL ksm( LL a,LL b, LL k)
10{
11 LL res = 1LL;
12 while (b)
13 {
14 if (b&1) res = (res * a) % k;
15 b = b >> 1LL;
16 a = ( a * a) % k;
17 }
18 return res;
19}
20int main()
21{
22 #ifndef ONLINE_JUDGE
23 freopen("code/in.txt","r",stdin);
24 #endif
1 while (~scanf("%d%d",&n,&m))
2 {
3 if (n==0&&m==0) break;
4 M.clear();
5 for ( int i = 1 ; i <= m ; i ++)
6 {
7 int s,t;
8 scanf("%d %d",&s,&t);
9 M.mat[s][t] = 1;
10 }
1 int T;
2 scanf("%d",&T);
3 while (T--)
4 {
5 int a,b,k;
6 scanf("%d%d%d",&a,&b,&k);
7 Mat ans;
8 ans.clear();
9 ans = M ^ k;
10 printf("%d\n",ans.mat[a][b]);
11 }
12 }
1 #ifndef ONLINE_JUDGE
2 fclose(stdin);
3 #endif
4 return 0;
5}