POJ 3268 Silver Cow Party（spfa）

Silver Cow Party

Language:
Silver Cow Party
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 30402 Accepted: 13798
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
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：给你n个点，m条边（单向边），目的地为X。求母牛们从各个点出发到X点聚会，之后再从X点返回原处，每个母牛都会走最短路径，问用时最长的母牛用时为多少。

题解：

正反建边跑最短路，求最大值。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
const int maxn=200000+10;
queue<int>q;
struct bb{
    int x,y,z;
}a[maxn];
struct cc{
    int from,to,cost;
}es[maxn];
int first[maxn],nxt[maxn];
bool vis[maxn];
int tot=0;
void build(int ff,int tt,int pp)
{
    es[++tot]=(cc){ff,tt,pp};
    nxt[tot]=first[ff];
    first[ff]=tot;
}
int dis1[maxn];
int dis2[maxn];
void spfa(int s)
{
    q.push(s);
    vis[s]=1;
    dis1[s]=0;
    while(!q.empty())
    {
        int u=q.front();
        q.pop();
        vis[u]=0;
        for(int i=first[u];i;i=nxt[i])
        {
            int v=es[i].to;
            if(dis1[v]>dis1[u]+es[i].cost)
            {
                dis1[v]=dis1[u]+es[i].cost;
                if(!vis[v])
                {
                    q.push(v);
                    vis[v]=1;
                }
            }
        }
    }
}
int main()
{
    int n,m,s;
    while(scanf("%d%d%d",&n,&m,&s)!=EOF)
    {
        tot=0;
        memset(first,0,sizeof(first));
        memset(nxt,0,sizeof(nxt));
        memset(dis1,63,sizeof(dis1));
        for(int i=1;i<=m;i++)
        {
            int x,y,z;
            scanf("%d%d%d",&a[i].x,&a[i].y,&a[i].z);
            build(a[i].x,a[i].y,a[i].z);
        }
        spfa(s);
        tot=0;
        memset(first,0,sizeof(first));
        memset(nxt,0,sizeof(nxt));
        for(int i=1;i<=m;i++)
        {
        	build(a[i].y,a[i].x,a[i].z);
        	dis2[i]=dis1[i];
        }
        memset(dis1,63,sizeof(dis1));
        spfa(s);
        int ans=0;
        for(int i=1;i<=n;i++)
        {
            ans=max(ans,dis1[i]+dis2[i]);
        }
        printf("%d\n",ans);
    }
    return 0;
}