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

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

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.

       這是一道有向圖最短路徑問題，題意是有n個點，m條有向邊，給定點x，求從其餘n-1個點到點x最短路徑的最大值

一開始用Dijkstra演算法，求出每個點到其餘n-1個點的最短路徑，時間複雜度為O(n^2),可惜TimeLimit,所以考慮另一種演算法

SPFA,時間複雜度O(n)

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define Inf 0x3f3f3f3f
const int N=1005;
int g[N][N];
int vis[N];
int dis[N][N];
int n,m,x;
void Init()
{
	memset(g,Inf,sizeof(g));
	for(int i=1;i<=n;i++)
	    g[i][i]=0;
}
void GetMap()
{
	for(int i=0;i<m;i++)
	{
		int u,v,d;
		cin>>u>>v>>d;
		if(g[u][v]>=d)
		  g[u][v]=d;
	}
}
void Dijkstra(int s)
{
	memset(vis,0,sizeof(vis));
	for(int i=1;i<=n;i++)
	    dis[s][i]=g[s][i];
	vis[s]=1;
	int u=s;
	for(int k=1;k<n;k++)
	{
		int minn=Inf;
		for(int i=1;i<=n;i++)
		   if(dis[s][i]<minn && !vis[i])
		   {
		   	   minn=dis[s][i];
		   	   u=i;
		   }
		vis[u]=1;
		for(int i=1;i<=n;i++)
		{
			if(dis[s][i]>dis[s][u]+g[u][i])
			     dis[s][i]=dis[s][u]+g[u][i];
		}
	}
}
int main()
{
	cin>>n>>m>>x;
	Init();
	GetMap();
	for(int i=1;i<=n;i++)
     	Dijkstra(i);
	int ans=0;
	for(int i=1;i<=n;i++)
	{
		if(i!=x)
		{
			ans=max(ans,dis[i][x]+dis[x][i]);
	    }   
		   
	}
	cout<<ans<<endl;
	return 0;
}

SPFA演算法