poj 3613Cow Relays

Description

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 ≤ I1i ≤ 1,000; 1 ≤ I2i ≤ 1,000), each of which is the termination for at least two trails. The cows know the lengthi of each trail (1 ≤ lengthi  ≤ 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: lengthi , I1i , and I2i

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
Source

USACO 2007 November Gold

题面

用一个矩阵a(i, j)来表示i到j经过若干条边的最短路，
初始化a为i到j边的长度，没有则是正无穷。
比如a矩阵表示经过n条边，b矩阵表示经过m条边，
那么a * b得到的矩阵表示经过m + n条边，
采用Floyd的思想进行更新。

 #include<iostream>
 #include<cstring>
 #include<cstdio>
 #include<algorithm>
 #include<cmath>
 #include<queue>
 #include<string>
 #include<map>
 #define ll long long
 using namespace std;
 ll n,m,S,T,l;
 map<ll,ll>id;
 struct node{
     ll a[][];
     friend node operator *(node x,node y)
     {
          node z;
          memset(z.a,0x3f,sizeof(z.a));
          for(ll k=;k<=l;k++)
           for(ll i=;i<=l;++i)
            for(ll j=;j<=l;++j)
             z.a[i][j]=min(z.a[i][j],x.a[i][k]+y.a[k][j]);
          return z;
     }
 }s,ans;
 void ksm()
 {
     ans=s;
     n--;
     while(n)
     {
         if(n&) ans=ans*s;
         s=s*s;
         n>>=;
     }
 }
 int main()
 {
     freopen("run.in","r",stdin);
     freopen("run.out","w",stdout);
     memset(s.a,0x3f,sizeof(s.a));
     scanf("%lld%lld%lld%lld",&n,&m,&S,&T);
     for(ll i=,x,y,z;i<=m;++i)
     {
        scanf("%lld%lld%lld",&z,&x,&y);
        if(id[x]) x=id[x];
        else l++,id[x]=l,x=l;
        if(id[y]) y=id[y];
        else l++,id[y]=l,y=l;
        s.a[x][y]=s.a[y][x]=z;
     }
     S=id[S];T=id[T];
     ksm();
     printf("%lld",ans.a[S][T]);
     return ;
 }
 /*
 2 6 6 4
 11 4 6
 4 4 8
 8 4 9
 6 6 8
 2 6 9
 3 8 9
 10
 */