Silver Cow Party

Time Limit: 2000MS	Memory Limit: 65536K
Total Submissions: 18401	Accepted: 8421

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. dijkstral演算法，同時求去的回來的最短路，相加後的最大值就是所求 程式碼：

#include<cstdio>
#include<cstring> 
#include<cmath>
#define INF 0x3f3f3f
#define MAXN 1000+10
int min(int x,int y)
{
	return x<y?x:y;
}
int max(int x,int y)
{
	return x>y?x:y;
} 
int n,m,p;
int a,b,c;
int cost[MAXN][MAXN];
int d[MAXN],vis[MAXN];//去的最短距離 
int d1[MAXN],vis1[MAXN];//回來的最短距離 
void djs(int s)
{
	memset(vis,0,sizeof(vis));
	memset(vis1,0,sizeof(vis1));
	memset(d,INF,sizeof(d));
	memset(d1,INF,sizeof(d1));
	d[s]=0;
	d1[s]=0;
	for(int i=1;i<=n;i++)//n個節點 
	{
		int v=-1;
		int v1=-1;
		//相鄰最小距離的頂點 
		for(int u=1;u<=n;u++)
		{
			if(!vis[u]&&(v==-1||d[u]<d[v]))
				v=u;
			if(!vis1[u]&&(v1==-1||d1[u]<d1[v1]))
				v1=u;
		}
		vis[v]=1;
		vis1[v1]=1;
		for(int u=1;u<=n;u++)
		{
			d[u]=min(d[u],d[v]+cost[v][u]);
			d1[u]=min(d1[u],d1[v1]+cost[u][v1]);
		 } 		 
	}
}
 int main()
 {
 	while(~scanf("%d%d%d",&n,&m,&p))
 	{
	 	memset(cost,INF,sizeof(cost));
	 	for(int i=1;i<=m;i++)
	 	{
	 		scanf("%d%d%d",&a,&b,&c);
	 		if(cost[a][b]>c)
	 			cost[a][b]=c;
		 }
		 djs(p);
		 int maxn=0;
		 for(int i=1;i<=n;i++)
		 {
		 	if(i==p)
		 		continue;
		 	maxn=max(maxn,d[i]+d1[i]);
		 }
		 printf("%d\n",maxn);
	 }
 	return 0;
 }