POJ 3268 Silver Cow Party （最短路）

Silver Cow Party

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 10147		Accepted: 4497

Description

One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the big cow party to be held at farm #X (1 ≤ X ≤ N). A total of M (1 ≤ M ≤ 100,000) unidirectional (one-way roads connects pairs of farms; road i requires T_i (1 ≤ T_i ≤ 100) units of time to traverse.

Each cow must walk to the party and, when the party is over, return to her farm. Each cow is lazy and thus picks an optimal route with the shortest time. A cow's return route might be different from her original route to the party since roads are one-way.

Of all the cows, what is the longest amount of time a cow must spend walking to the party and back?

Input

Line 1: Three space-separated integers, respectively:  N,  M, and  X 
Lines 2.. M+1: Line  i+1 describes road  i with three space-separated integers:  A_i,  B_i, and  T_i. The described road runs from farm  A_i to farm  B_i, requiring  T_i time units to traverse.

Output

Line 1: One integer: the maximum of time any one cow must walk.

Sample Input

4 8 2
1 2 4
1 3 2
1 4 7
2 1 1
2 3 5
3 1 2
3 4 4
4 2 3

Sample Output

10

Hint

Cow 4 proceeds directly to the party (3 units) and returns via farms 1 and 3 (7 units), for a total of 10 time units.

Source

USACO 2007 February Silver

 

 

 

练习模板用的。

用原图和逆图分别用一次单源最短路。

 

Dijkstra算法

//============================================================================
// Name        : POJ.cpp
// Author      : 
// Version     :
// Copyright   : Your copyright notice
// Description : Hello World in C++, Ansi-style
//============================================================================

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
const int MAXN=1010;
const int INF=0x3f3f3f3f;
bool vis[MAXN];
void Dijkstra(int cost[][MAXN],int lowcost[],int n,int beg)
{
    for(int i=1;i<=n;i++)
    {
        lowcost[i]=INF;
        vis[i]=false;
    }
    lowcost[beg]=0;
    for(int j=0;j<n;j++)
    {
        int k=-1;
        int Min=INF;
        for(int i=1;i<=n;i++)
            if(!vis[i]&&lowcost[i]<Min)
            {
                Min=lowcost[i];
                k=i;
            }
        if(k==-1)break;
        vis[k]=true;
        for(int i=1;i<=n;i++)
            if(!vis[i]&&lowcost[k]+cost[k][i]<lowcost[i])
                lowcost[i]=lowcost[k]+cost[k][i];
    }
}
int dist1[MAXN];
int dist2[MAXN];
int cost[MAXN][MAXN];
int main()
{
//    freopen("in.txt","r",stdin);
//    freopen("out.txt","w",stdout);
    int N,M,X;
    int u,v,w;
    while(scanf("%d%d%d",&N,&M,&X)==3)
    {
        for(int i=1;i<=N;i++)
            for(int j=1;j<=N;j++)
            {
                if(i==j)cost[i][j]=0;
                else cost[i][j]=INF;
            }
        while(M--)
        {
            scanf("%d%d%d",&u,&v,&w);
            cost[u][v]=min(cost[u][v],w);
        }
        Dijkstra(cost,dist1,N,X);
        for(int i=1;i<=N;i++)
            for(int j=1;j<i;j++)
                swap(cost[i][j],cost[j][i]);
        Dijkstra(cost,dist2,N,X);
        int ans=0;
        for(int i=1;i<=N;i++)
            ans=max(ans,dist1[i]+dist2[i]);
        printf("%d\n",ans);
    }
    return 0;
}