/*
Silver Cow Party
Time Limit: 2000MS  Memory Limit: 65536K 
Total Submissions: 3321  Accepted: 1407 

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 Ti (1 ≤ Ti ≤ 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: Ai, Bi, and Ti. 
The described road runs from farm Ai to farm Bi, requiring Ti 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
*/
#define MAXN 1000
#include<iostream>
#include<memory.h>
using namespace std;
class edge{
	public:
		int to;
	    int cost;
		edge *next;
		edge(int _to,int _cost,edge *_next):to(_to),cost(_cost),next(_next){}
} *hLink[MAXN+1];
int n,m,x;
int mincost[MAXN+1],cost[MAXN+1];
int Queue[MAXN*MAXN],qH,qT;
bool inQueue[MAXN+1];
void spfa(int s){
	memset(inQueue,false,sizeof(inQueue));
	for (int i=1;i<=n;++i) cost[i]=0x7fff;
	cost[s]=0;inQueue[s]=true;
	Queue[qH=0]=s;qT=1;
	while (qH!=qT){
		int cur=Queue[qH++];qH%=MAXN*MAXN;
		inQueue[cur]=false;
		for (edge *p=hLink[cur];p;p=p->next)
			if (cost[p->to]>cost[cur]+p->cost){
			   cost[p->to]=cost[cur]+p->cost;
			   if (!inQueue[p->to]){
				   inQueue[p->to]=true;
				   Queue[qT++]=p->to;
				   qT%=MAXN*MAXN;
			   }
			}
	}
}
int main (void){
	int a,b,c;
	cin>>n>>m>>x;
	for (int i=0;i<m;++i){
		cin>>a>>b>>c;
		hLink[a]=new edge(b,c,hLink[a]);
	}
	for (int i=1;i<=n;++i){
		spfa(i);
		if (i!=x) mincost[i]+=cost[x];
		else{
			for (int j=1;j<=n;++j)
				if (j!=x) mincost[j]+=cost[j];
		}
	}
	int maximum=0;
	for (int i=1;i<=n;++i)
		if (i!=x) maximum=max(maximum,mincost[i]);
	cout<<maximum<<endl;
	return 0;
}

• Re:SPFA 360ms,ASK FOR 0ms Solution (10) fanlonglong8500 2009-08-31 12:08:11