POJ 3268 Silver Cow Party（取反最短路）

思路：一个有向图要求来回的最短路，一个经典的做法是保存一个原图，一个边取反的图，然后跑两遍spfa就可以了

#include
#include
#include
#include
#include
using namespace std;
const int maxn = 1005;
#define inf 1e9
vector >e1[maxn];
vector >e2[maxn];
int d1[maxn],d2[maxn];
int vis[maxn];
int n,m,s;
void spfa(vector >*e,int d[])
{
	memset(vis,0,sizeof(vis));
	queueq;
	for(int i = 0;i<=n;i++)d[i]=inf;
	d[s]=0;
	q.push(s);
	while(!q.empty())
	{
		int u = q.front();q.pop();
		vis[u]=0;
		for(int i = 0;id[u]+w)
			{
				d[v]=d[u]+w;
				if(!vis[v])
				{
					vis[v]=1;
					q.push(v);
				}
			}
		}
	}
}
int main()
{
	scanf("%d%d%d",&n,&m,&s);
	for(int i = 0;i

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.