Cow Relays(Floyed+倍增)

题目描述

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.

 

输入

* 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

 

输出

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

 

样例输入

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

 

样例输出

10
倍增的思想，通过矩阵快速幂进行优化。

#include <bits/stdc++.h>
#define maxn 105
using namespace std;
typedef long long ll;
int cnt=1;
struct Matrix{
    int a[maxn][maxn];
    Matrix operator * (const Matrix &x) const
    {
        Matrix c;
        memset(c.a,0x3f,sizeof(c.a));
        for (int k=1; k<=cnt; k++)
            for (int i=1; i<=cnt; i++)
                for (int j=1; j<=cnt; j++)
                    c.a[i][j] = min(c.a[i][j],a[i][k]+x.a[k][j]);
        return c;
     }
}s,ans;
void ksm(int n)
{
    ans = s;
    n--;
    for (; n; n>>=1)
    {
        if (n&1) ans = ans*s;
        s = s*s;
    }
}
int vis[1005];
int main()
{
   int n,m,S,t;
   //freopen("in.txt","r",stdin);
   scanf("%d%d%d%d",&n,&m,&S,&t);
   memset(s.a,0x3f,sizeof(s.a));
   for(int i=1;i<=m;i++)
   {
       int x,u,v;
       scanf("%d%d%d",&x,&u,&v);
       if(!vis[u])
       {
           vis[u]=++cnt;
       }
       if(!vis[v])
       {
           vis[v]=++cnt;
       }
       s.a[vis[u]][vis[v]]=s.a[vis[v]][vis[u]]=x;
   }
   ksm(n);
   printf("%d\n",ans.a[vis[S]][vis[t]]);
   return 0;
}