【图论补完计划】poj 3613（Floyd 快速幂）

Cow Relays

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7463		Accepted: 2922

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

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <set>

using namespace std;

const int maxn=1005;
const int inf=1e9;

int mp[maxn][maxn];
int tmp[maxn][maxn];
int mmp[maxn][maxn];
int res[maxn][maxn];
int cord[maxn];
bool used[maxn];

int n,m,st,ed;
int num;

void init(){
    for(int i=0;i<maxn;i++){
        for(int j=0;j<maxn;j++){
            mp[i][j]=inf;
            tmp[i][j]=inf;
            mmp[i][j]=inf;
            res[i][j]=inf;
        }
        res[i][i]=0;
    }
}

void floyd(int a[][maxn],int b[][maxn],int c[][maxn]){
    for(int k=0;k<num;k++)
        for(int i=0;i<num;i++)
            for(int j=0;j<num;j++)
                if(a[cord[i]][cord[j]]>b[cord[i]][cord[k]]+c[cord[k]][cord[j]])
                    a[cord[i]][cord[j]]=b[cord[i]][cord[k]]+c[cord[k]][cord[j]];
}

void trans(int a[][maxn],int b[][maxn]){
    for(int i=0;i<num;i++){
        for(int j=0;j<num;j++){
            a[cord[i]][cord[j]]=b[cord[i]][cord[j]];
            b[cord[i]][cord[j]]=inf;
        }
    }
}

void quick_mod(){
    while(n){
        if(n&1){
            floyd(mmp,res,mp);
            trans(res,mmp);
        }
        floyd(tmp,mp,mp);
        trans(mp,tmp);
        n>>=1;
    }
}

int main(){
    scanf("%d %d %d %d",&n,&m,&st,&ed);
    init();
    for(int i=0;i<m;i++){
        int u,v,w;
        scanf("%d %d %d",&w,&u,&v);
        if(!used[u]){
            used[u]=1;
            cord[num++]=u;
        }
        if(!used[v]){
            used[v]=1;
            cord[num++]=v;
        }
        mp[u][v]=mp[v][u]=min(mp[u][v],w);
    }
    quick_mod();
    printf("%d
",res[st][ed]);
    return 0;
}