Silver Cow Party POJ - 3268（（正+反）最短路）

Silver Cow Party

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 27217		Accepted: 12436

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

Sample Output

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：n个农场的牛打算去编号为x的农场开一个聚会，并且，最后要回到原来的农场。农场与农场之间通过有向的路联通（保证任意农场必定联通）现在，每条路都有一个权值，牛总是走的路总是保证权值最小。问这些牛中，走过的路的权值最大的是多少？

思路：两个最短路即可。牛往回走，其实就相当与把原来的有向图的边全部反过来！！明白了这点就很容易做了

#include "iostream"
#include "queue"
#include "algorithm"
#include "cstring"
using namespace std;
const int Max=1005;
const int inf=0xfffffff;
int G1[Max][Max],G2[Max][Max];//分别正向建边，和反向建边
int d1[Max],d2[Max];//分别记录正，反向建边，从农场x到其他农场的最权值
int n,e,s;
struct edge
{
    int to,cost;
    friend bool operator <(edge a,edge b){return a.cost<b.cost;}
};
void dijk()
{
    priority_queue<edge> que1;
    priority_queue<edge> que2;
    fill(d1,d1+n+1,inf);
    fill(d2,d2+n+1,inf);
    d1[s]=d2[s]=0;
    edge e1,e2;
    e1.to=e2.to=s;
    e1.cost=e2.cost=0;
    que1.push(e1);
    que2.push(e2);
    while(!que1.empty()){//正向最短路
        e1=que1.top();
        que1.pop();
        int u=e1.to;
        for(int i=1;i<=n;i++){
            int v=G1[u][i];
            if(v!=0&&d1[i]>d1[u]+v){
                d1[i]=d1[u]+v;
                e1.to=i;
                e1.cost=d1[i];
                que1.push(e1);
            }
        }
    }
    while(!que2.empty()){//逆向最短路
        e2=que2.top();
        que2.pop();
        int u=e2.to;
        for(int i=1;i<=n;i++){
            int v=G2[u][i];
            if(v!=0&&d2[i]>d2[u]+v){
                d2[i]=d2[u]+v;
                e2.to=i;
                e2.cost=d2[i];
                que2.push(e2);
            }
        }
    }
}
int main()
{
    int u,v,w;
    ios::sync_with_stdio(false);
    while(cin>>n>>e>>s){
        memset(G1,0, sizeof(G1));
        memset(G2,0, sizeof(G2));
        for(int i=0;i<e;i++){
            cin>>u>>v>>w;
            G1[u][v]=w;
            G2[v][u]=w;
        }
        dijk();
        int ans=0;
        for(int i=1;i<=n;i++){
            ans=max(ans,d1[i]+d2[i]);
        }
        cout<<ans<<endl;
    }
}

Silver Cow Party POJ - 3268（（正+反）最短路）

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