poj 3613 Cow Relays

Cow Relays

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5411		Accepted: 2153

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 ≤ 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

n次floyd，原来flody也能够像矩阵一样高速幂。详细的能够看论文《矩阵乘法在信息学的应用》

代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
const int inf=(1<<30)-1;
int n;
int hash[3010];//映射
struct matrix
{
    int ma[210][210];
    matrix()
    {
       for(int i=0;i<210;i++)
          for(int j=0;j<210;j++)
           ma[i][j]=inf;
    }
};
matrix floyd(matrix x,matrix y)
{
    matrix ans;
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=n;j++)
        {
            for(int k=1;k<=n;k++)
            ans.ma[i][j]=min(ans.ma[i][j],x.ma[i][k]+y.ma[k][j]);
        }
    }
    return ans;
}
matrix pow(matrix a,int m)
{
    matrix ans;
    for(int i=1;i<=n;i++)
    ans.ma[i][i]=0;
    while(m)
    {
        if(m&1)
        ans=floyd(ans,a);
        a=floyd(a,a);
        m=m>>1;
    }
    return ans;
}
int main()
{
    int k,t,s,e;
    while(~scanf("%d%d%d%d",&k,&t,&s,&e))
    {
        int x,y,d;
        memset(hash,0,sizeof(hash));
        n=1;
        matrix a;
        for(int i=0;i<t;i++)
        {
            scanf("%d%d%d",&d,&x,&y);
            if(!hash[x])
            hash[x]=n++;
            if(!hash[y])
            hash[y]=n++;
            a.ma[hash[x]][hash[y]]=a.ma[hash[y]][hash[x]]=d;
        }
        n=n-1;
        matrix ans;
        ans=pow(a,k);
        printf("%d\n",ans.ma[hash[s]][hash[e]]);
    }
    return 0;
}