Cow Relays POJ - 3613 （floyd+快速幂）

For their physical fitness program, N (2 ≤ N ≤ 1,000,000) cows have decided to run a relay race using the T (2 ≤ T ≤ 100) cow trails throughout the pasture.

Each trail connects two different intersections (1 ≤ I_1i ≤ 1,000; 1 ≤ I_2i ≤ 1,000), each of which is the termination for at least two trails. The cows know the length_i of each trail (1 ≤ length_i  ≤ 1,000), the two intersections the trail connects, and they know that no two intersections are directly connected by two different trails. The trails form a structure known mathematically as a graph.

To run the relay, the N cows position themselves at various intersections (some intersections might have more than one cow). They must position themselves properly so that they can hand off the baton cow-by-cow and end up at the proper finishing place.

Write a program to help position the cows. Find the shortest path that connects the starting intersection (S) and the ending intersection (E) and traverses exactly N cow trails.

Input

* Line 1: Four space-separated integers: N, T, S, and E
* Lines 2..T+1: Line i+1 describes trail i with three space-separated integers: length_i , I_1i , and I_2i

Output

* Line 1: A single integer that is the shortest distance from intersection S to intersection E that traverses exactly N cow trails.

Sample Input

2 6 6 4
11 4 6
4 4 8
8 4 9
6 6 8
2 6 9
3 8 9

Sample Output

10

题意：求过s，t两点的刚好经过k条边的最短路
思路：任意最短路可以想到Floyd，（01矩阵的乘积A^k中，A[i][j]代表刚好经过k条边的从i到j的数量）
maps【i】【j】 为 经过一条边的最短路， 对于maps【i】【k】 + maps【k】【j】 可以看出是经过两条边的最短路
那么对于
r+m == k 且 A为经过r条边的最短路，B为经过m条边的最短路，通过maps【i】【k】+maps【k】【j】就得到了刚好经过k条边的最短路

 #include<cstdio>
 #include<iostream>
 #include<cstring>
 using namespace std;

 int n,k,m,s,t;
 int has[];
 struct matrix
 {
     int maps[][];
     matrix operator *(const matrix &x)const
     {
         matrix c;
         memset(c.maps,0x3f,sizeof(c.maps));
         for(int k=;k<=n;k++)
         {
             for(int i=;i<=n;i++)
             {
                 for(int j=;j<=n;j++)
                 {
                     c.maps[i][j] = min(c.maps[i][j],maps[i][k]+x.maps[k][j]);
                 }
             }
         }
         return c;
     }
 };

 matrix qpow(matrix a,int k)
 {
     matrix ans = a;
     k--;
     while(k)
     {
         if(k&)ans = ans * a;
         a = a*a;
         k >>= ;
     }
     return ans;
 }
 int main()
 {
     while(~scanf("%d%d%d%d",&k,&m,&s,&t))
     {
         int tot=;
         matrix ans;
         memset(ans.maps,0x3f,sizeof(ans.maps));
         for(int i=;i<=m;i++)
         {
             int w,x,y;
             scanf("%d%d%d",&w,&x,&y);
             if(!has[x])
             {
                 has[x] = ++tot;
             }
             if(!has[y])
             {
                 has[y] = ++tot;
             }
             if(w < ans.maps[has[x]][has[y]])
             {
                 ans.maps[has[x]][has[y]] = ans.maps[has[y]][has[x]] = w;
             }
         }
         n = tot;
         ans = qpow(ans,k);
         printf("%d\n",ans.maps[has[s]][has[t]]);
     }
 }