POJ 3268 Dijkstra+priority

//两个Dijkstra，虽然是多元最短路径，但是不能
//用Floyd，会TLE，模板题
#include<iostream>
#include<queue>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxv=1010;
const int maxe=100010;
const int INF=0x3f3f3f3f;
int head[maxv],dis[maxv],cnt,cnt2,head2[maxv];
bool visit[maxv];
struct Edge{
    int to,val,next;
}edge[maxe],edge2[maxe];
struct Node{
    int u,dis;
    bool operator< (const Node rhs) const{
        return this->dis>rhs.dis;
    }
}now,temp;
void init(){
    memset(head,-1,sizeof head);
    memset(head2,-1,sizeof head2);
    memset(dis,INF,sizeof dis);
}
void addedge(int from,int to,int val){
    edge[cnt].to=to;
    edge[cnt].val=val;
    edge[cnt].next=head[from];
    head[from]=cnt++;
}
void addedge2(int from,int to,int val){
    edge2[cnt2].to=to;
    edge2[cnt2].val=val;
    edge2[cnt2].next=head2[from];
    head2[from]=cnt2++;
}
void Dijkstra(int s,int head[],Edge edge[],int dist[],bool visit[]) {
    int i,v;
    dist[s]=0;
    priority_queue<Node> pq;
    temp.dis=0,temp.u=s;
    pq.push(temp);
    while(!pq.empty()) {
        temp=pq.top();
        pq.pop();
        if(visit[temp.u]==true)
            continue;
        visit[temp.u]=true;
        for(i=head[temp.u]; i!=-1; i=edge[i].next) {
            v=edge[i].to;
            if(visit[v]==false&&dist[v]>dist[temp.u]+edge[i].val) {
                dist[v]=dist[temp.u]+edge[i].val;
                now.u=v;
                now.dis=dist[v];
                pq.push(now);
            }
        }
    }
}
int main(void) {
#ifndef ONLINE_JUDGE
    freopen("E:\\input.txt","r",stdin);
#endif // ONLINE_JUDGE
    int v,e,x,ans;
    init();
    scanf("%d%d%d",&v,&e,&x);
    int t1,t2,t3;
    for(int i=0;i<e;i++){
        scanf("%d%d%d",&t1,&t2,&t3);
        addedge(t1,t2,t3);
        addedge2(t2,t1,t3);
    }
    Dijkstra(x,head2,edge2,dis,visit);//去程
    int ret[maxv]={0};
    for(int i=1;i<=v;i++){
        ret[i]+=dis[i];
    }
    memset(dis,INF,sizeof dis);
    memset(visit,false ,sizeof visit);
    Dijkstra(x,head,edge,dis,visit);//返程
    for(int i=1;i<=v;i++)
        ret[i]+=dis[i];
    ans=(*max_element(ret+1,ret+v+1));
    printf("%d\n",ans);
    return 0;
}