POJ-3268-最短路(dijkstra算法)

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?

#include<queue>
#include<vector>
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
struct Edge
{
    int to;
    int dis;
    Edge(int to, int dis){
        this -> to = to;
        this -> dis = dis;
    }
};
typedef pair<int,int>P;

int a,b,c;
int N,M,X;
int d1[1005],d2[1005];
vector<Edge> G1[1005];
vector<Edge> G2[1005];
void dijkstra(int s,int d[],vector<Edge> G[])
{
    priority_queue<P,vector<P>,greater<P> >q;
    d[s]=0;
    q.push(P(0,s));
    while(q.size())
    {
        P p=q.top();
        q.pop();
        int v=p.second;
        for(int i=0;i<G[v].size();i++)
        {
            Edge& e=G[v][i];
            if(d[e.to]>d[v]+e.dis)
            {
                d[e.to]=d[v]+e.dis;
                q.push(P(d[e.to],e.to));
            }
        }
    }
}
int main()
{
    memset(d1,0x5f,sizeof(d1));
    memset(d2,0x5f,sizeof(d2));
    scanf("%d%d%d",&N,&M,&X);
    for(int i=1;i<=M;i++)
    {
        scanf("%d%d%d",&a,&b,&c);
        G1[a].push_back(Edge(b,c));
        G2[b].push_back(Edge(a,c));
    }
    dijkstra(X,d1,G1);
    dijkstra(X,d2,G2);
    int small_max=-1;
    for(int i=1;i<=N;i++)
    {
        if(i==X) continue;
        small_max=max(small_max,d1[i]+d2[i]);
    }
    cout<<small_max<<endl;
}