POJ3268————Silver Cow Party（最短路）

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本文介绍了一种解决特定场景下的最短路径问题的算法实现，该问题涉及多个农场间的单向道路连接，每头牛从各自所在的农场出发参加聚会再返回，目标是最小化行走时间。通过使用Dijkstra算法进行求解。

Silver Cow Party

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 19688		Accepted: 9008

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

Sample Output

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.

裸的最短路

熟悉一下

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
using namespace std;
const int MAXN=1010;
const int MAXM=100010;
const int INF=1000000007;
int tot=0;
int head[MAXN];
struct edge{
	int v,w,next;
};
edge e[MAXM];
void add(int u,int v,int w){e[tot].v=v;e[tot].w=w;e[tot].next=head[u];head[u]=tot++;}
int d[MAXN][MAXN];
typedef pair <int,int>    P;  
void dij(int s){  
    priority_queue<P,vector<P>,greater<P> > que;  
    fill(d[s],d[s]+MAXN,INF);  
    d[s][s]=0;  
    que.push(P(0,s));  
    while(!que.empty()){  
        P p=que.top();  
        que.pop();  
        int u=p.second;  
        if(d[s][u]<p.first)  
            continue;  
        for(int k=head[u];k!=-1;k=e[k].next){  
            int v=e[k].v;  
            int w=e[k].w;   
            if(d[s][v]>d[s][u]+w){  
                d[s][v]=d[s][u]+w;  
                que.push(P(d[s][v],v));  
            }  
        }  
    }  
}  
int main()
{
	int n,m,x;
	scanf("%d%d%d",&n,&m,&x);
	tot=0;
	memset(head,-1,sizeof(head));
	for(int i=0;i<m;i++){
		int u,v,w;
		scanf("%d%d%d",&u,&v,&w);
		add(u,v,w);
	}
	for(int i=1;i<=n;i++){
		dij(i);
	}
	int MAX=0;
	for(int i=1;i<=n;i++){
		MAX=max(MAX,d[i][x]+d[x][i]);
	}
	printf("%d\n",MAX);
}