第一次理解错成每只母牛往返的必须时间，所以求最小，事实上并不是。
还有看了讨论，居然有矩阵转置的解法，学习一下拓展思维。
#include <stdio.h>
#include <memory.h>
#include <queue>
#define MAXN 1001
#define INF 0x3f3f3f3f
using namespace std;
typedef struct
{	int start,end,time;
	int next1,next2;	
}Edge;
Edge de[100001];
int headlist1[MAXN],headlist2[MAXN];
int dis1[MAXN],dis2[MAXN],visited[MAXN];
void spfa1(int k)
{	queue<int>q;
	int i,temp,x,y;
	memset(visited,0,sizeof(visited));
	memset(dis1,0x3f,sizeof(dis1));
	q.push(k),visited[k]=1,dis1[k]=0;
	while(!q.empty())
	{	temp=q.front();
		q.pop();
		visited[temp]=0;
		for(i=headlist1[temp];i!=-1;i=de[i].next1)
		{	x=de[i].start,y=de[i].end;
			if(dis1[y]>dis1[x]+de[i].time)
			{	dis1[y]=dis1[x]+de[i].time;
				if(!visited[y])
					visited[y]=1,q.push(y);
			}
		} 	
	}
}
void spfa2(int k)
{	queue<int>q;
	int i,temp,x,y;
	memset(visited,0,sizeof(visited));
	memset(dis2,0x3f,sizeof(dis2));
	q.push(k),visited[k]=1,dis2[k]=0;
	while(!q.empty())
	{	temp=q.front();
		q.pop();
		visited[temp]=0;
		for(i=headlist2[temp];i!=-1;i=de[i].next2)
		{	x=de[i].end,y=de[i].start;
			if(dis2[y]>dis2[x]+de[i].time)
			{	dis2[y]=dis2[x]+de[i].time;
				if(!visited[y])
					visited[y]=1,q.push(y);
			}
		} 	
	}
}
int main()
{	int N,M,X,i;
	while(scanf("%d %d %d",&N,&M,&X)!=EOF)
	{	memset(headlist1,-1,sizeof(headlist1));
		memset(headlist2,-1,sizeof(headlist2));
		for(i=0;i<M;++i)
		{	scanf("%d %d %d",&de[i].start,&de[i].end,&de[i].time);
			de[i].next1=headlist1[de[i].start];
			headlist1[de[i].start]=i;
			de[i].next2=headlist2[de[i].end];
			headlist2[de[i].end]=i;
		}
		spfa1(X);
		spfa2(X);
		int max=-1;
		for(i=1;i<=N;++i)
			if(max<dis1[i]+dis2[i]&&i!=X)
				max=dis1[i]+dis2[i];			
		printf("%d\n",max);
	}
	return 0;
}