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.

題意：有n個農廠，n頭牛，其中標號為x的農場要舉行party,其餘農場的牛要去參加，但是牛都很懶，所以不管是過去還是回來，牛都會走對於他們自己而言最短的路，要你求出所有牛中走得時間最長的一個。

分析：這道題首先計算出每頭牛來回的最短距離和，然後再進行比較，輸出最大值。

我的方法是用兩次Dijkstra，分別算出去和回來的最小值，再求和。

#include<stdio.h>
#include<string.h>

int m,n,u,max,e[1005][1005],dis1[1005],dis2[1005],book[1005];
int inf = 99999999;
void Dijkstra()
{
	int i,j,k,min;
	for(i = 1; i <= n; i ++)
	{
		dis1[i] = e[u][i];
		dis2[i] = e[i][u];
		book[i] = 0;
	}
	for(i = 1; i < n; i ++)
	{
		min = inf;
		for(j = 1; j <= n; j ++)
			if(book[j] == 0 && dis1[j] < min)
			{
				min = dis1[j];
				k = j; 
			}
		book[k] = 1;
		for(j = 1; j <= n; j ++)
			if(book[j] == 0 && dis1[j] > dis1[k] + e[k][j])
				dis1[j] = dis1[k] + e[k][j];
	}
	memset(book,0,sizeof(book));
	for(i = 1; i < n; i ++)
	{
		min = inf;
		for(j = 1; j <= n; j ++)
			if(book[j] == 0 && dis2[j] < min)
			{
				min = dis2[j];
				k = j; 
			}
		book[k] = 1;
		for(j = 1; j <= n; j ++)
			if(book[j] == 0 && dis2[j] > dis2[k] + e[j][k])
				dis2[j] = dis2[k] + e[j][k];
	}
	for(i = 1; i <= n; i ++)
	{
		if(dis1[i] + dis2[i] > max)
			max = dis1[i] + dis2[i];
	}
	printf("%d\n",max);
}
int main()
{
	int i,j,t1,t2,t3;
	while(scanf("%d%d%d",&n,&m,&u) != EOF)
	{
		max = 0;
		for(i = 1; i <= n; i ++)
			for(j = 1; j <= n; j ++)
				if(i == j)
					e[i][j] = 0;
				else
					e[i][j] = inf;
		for(i = 1; i <= m; i ++)
		{
			scanf("%d%d%d",&t1,&t2,&t3);
			e[t1][t2] = t3;
		}
		Dijkstra();
	}
	return 0;
}