poj 3268 Silver Cow Party（最短路）

Silver Cow Party

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 17017		Accepted: 7767

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.

 

题意：有n个牛分别住在n个农场，现在要在x农场办party，每个农场的牛都要去参加，所有的牛在去和回来的时候都会选择花费时间最短的路线，现在问，在所有的牛中花费时间最长的是多少时间（注意路径是单向的所以去和回来的路可能不同）

题解：每头牛都起点到终点，终点到起点跑两次最短路算法，求出最大的

#include<stdio.h>
#include<string.h>
#include<queue>
#include<cstdio> 
#include<string>
#include<math.h>
#include<algorithm>
#define LL long long
#define PI atan(1.0)*4
#define DD double
#define MAX 110000
#define mod 100
#define dian 1.000000011
#define INF 0x3f3f3f
using namespace std;
int head[MAX],ans;
int n,m,j,i,t,k,x;
int a,b,c;
int vis[MAX],dis[MAX];
struct node 
{
	int u,v,w;
	int next;
}edge[MAX];
void add(int u,int v,int w)
{
	edge[ans].u=u;
	edge[ans].v=v;
	edge[ans].w=w;
	edge[ans].next=head[u];
	head[u]=ans++;
}
void getmap()
{
	memset(head,-1,sizeof(head));
	ans=0;
	for(i=1;i<=m;i++)
	{
		scanf("%d%d%d",&a,&b,&c);
		add(a,b,c);
	}
}
int spfa(int sx,int sy)
{
    int i,j;
    queue<int>q;
    memset(vis,0,sizeof(vis));
    for(i=1;i<=n;i++)
        dis[i]=INF;
    vis[sx]=1;
    dis[sx]=0;
    q.push(sx);
    while(!q.empty())
    {
        int u=q.front();
            q.pop();
        vis[u]=0;
        for(i=head[u];i!=-1;i=edge[i].next)
        {
            int top=edge[i].v;
            if(dis[top]>dis[u]+edge[i].w)
            {
                dis[top]=dis[u]+edge[i].w;
                if(!vis[top])
                {
                    vis[top]=1;
                    q.push(top);
                }
            }
        }
    }
    return dis[sy];
}

void solve()
{
	int Max=-INF;
	int Min=INF;
	int ant;
	for(i=1;i<=n;i++)
	{
		ant=spfa(i,x)+spfa(x,i);
		Max=max(Max,ant);
	}
	printf("%d
",Max);
} 
int main()
{
	while(scanf("%d%d%d",&n,&m,&x)!=EOF)
	{
		getmap();
		solve();
	}
	return 0;
}