（poj）3268 Silver Cow Party 最短路

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 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

题目网址：http://poj.org/problem?id=3268

题意：有N只牛，编号从1到N他们从自己的地方到X去开会然后再回来，他们都选择最短的路径，问从去到回来，每只牛走的最远距离是多少？

方法：先求从X到各个点的最短路，然后把路径交换一下，再求一次从x到各点的最短路

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <stack>
#include <queue>
#include <math.h>
#include <vector>
using namespace std;
#define N  1010
#define ll long long
#define INF 0x3f3f3f3f
#define met(a,b) memset(a,b,sizeof(a));
vector<vector<int> >Q;
struct node
{
    int u,l,next;
}s[200020];
int a[N],b[N],k,n;
int vis[N],used[N],dis1[N],dis[N];
void add(int e,int f,int l)
{
    s[k].u=f;
    s[k].l=l;
    s[k].next=a[e];
    a[e]=k++;
    s[k].u=e;
    s[k].l=l;
    s[k].next=b[f];
    b[f]=k++;
}
void spfa1(int x)
{
    met(vis,0);met(used,0);
    for(int i=1;i<=n;i++)
        dis1[i]=INF;
    queue<int>q;
    int p=x,v;
    q.push(p);
    vis[x]=1;
    dis1[x]=0;
    used[x]=1;
    while(q.size())
    {
        p=q.front();
        vis[p]=0;
        q.pop();
        for(int i=a[p];i!=-1;i=s[i].next)
        {
            v=s[i].u;
            if(dis1[v]>dis1[p]+s[i].l)
            {
                dis1[v]=dis1[p]+s[i].l;
                q.push(v);
                vis[v]=1;
            }

        }

    }
}
void spfa2(int x)
{
    met(vis,0);met(used,0);
    for(int i=1;i<=n;i++)
        dis[i]=INF;
    queue<int>q;
    int p=x,v;
    q.push(p);
    vis[x]=1;
    dis[x]=0;
    used[x]=1;
    while(q.size())
    {
        p=q.front();
        vis[p]=0;
        q.pop();
        for(int i=b[p];i!=-1;i=s[i].next)
        {
            v=s[i].u;
            if(dis[v]>dis[p]+s[i].l)
            {
                dis[v]=dis[p]+s[i].l;
                q.push(v);
                vis[v]=1;
            }

        }

    }
}
int main()
{
    int m,x,e,f,l;
    while(scanf("%d %d %d",&n,&m,&x)!=EOF)
    {
        k=0;
        met(a,-1);met(b,-1);
        for(int i=0;i<m;i++)
        {
            scanf("%d %d %d",&e,&f,&l);
            add(e,f,l);
        }
        //int ans=spfa1();
        int ans=0;
        spfa1(x);
          spfa2(x);
        for(int i=1;i<=n;i++)
        {
            ans=max(ans,dis[i]+dis1[i]);
        }

        printf("%d\n",ans);
    }
    return 0;
}