PAT甲题题解-1030. Travel Plan (30)-最短路+输出路径

模板题
最短路+输出路径
如果最短路不唯一，输出cost最小的

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string.h>
#include <cmath>
#include <queue>
#define INF 0x3f3f3f3f
using namespace std;
const int maxn=505;
int n,m,s,t;

struct Edge{
    int to;
    int next;
    int dis;
    int cost;
}edge[maxn*maxn];
int head[maxn];
int tot;
void init(){
    memset(head,-1,sizeof(head));
    tot=0;
}
void add(int u,int v,int d,int cc){
    edge[tot].to=v;
    edge[tot].dis=d;
    edge[tot].cost=cc;
    edge[tot].next=head[u];
    head[u]=tot++;
}

struct Node{
    int u;
    int dis;
    int cost;
    bool operator<(const Node tmp)const{
        if(dis==tmp.dis){
            return cost>tmp.cost;
        }
        else{
            return dis>tmp.dis;
        }
    }
};

int dis[maxn];
int pre[maxn];
int vis[maxn];
int costs[maxn];
void dijkstra(int s,int t){
    for(int i=0;i<n;i++){
        dis[i]=INF;
        costs[i]=INF;
        pre[i]=-1;
        vis[i]=0;
    }
    priority_queue<Node>q;
    Node tmp,node;
    tmp.u=s;
    tmp.dis=0;
    tmp.cost=0;
    dis[s]=costs[s]=0;
    q.push(tmp);
    while(!q.empty()){
        tmp=q.top();
        q.pop();
        int u=tmp.u;
        vis[u]=1;
        for(int k=head[u];k!=-1;k=edge[k].next){
            int v=edge[k].to;
            if(!vis[v]){
                if(tmp.dis+edge[k].dis<dis[v]){
                    dis[v]=tmp.dis+edge[k].dis;
                    costs[v]=tmp.cost+edge[k].cost;
                    node.u=v;
                    node.dis=dis[v];
                    node.cost=costs[v];
                    q.push(node);
                    pre[v]=u;
                }
                else if(tmp.dis+edge[k].dis==dis[v] && tmp.cost+edge[k].cost<costs[v]){
                    costs[v]=tmp.cost+edge[k].cost;
                    node.u=v;
                    node.dis=dis[v];
                    node.cost=costs[v];
                    q.push(node);
                    pre[v]=u;
                }
            }
        }
    }
}
int main()
{
    scanf("%d %d %d %d",&n,&m,&s,&t);
    int u,v,d,cc;
    init();
    for(int i=0;i<m;i++){
        scanf("%d %d %d %d",&u,&v,&d,&cc);
        add(u,v,d,cc);
        add(v,u,d,cc);
    }
    dijkstra(s,t);
    int route[maxn];
    int cnt=0;
    int id=t;
    route[cnt++]=t;
    while(pre[id]!=-1){
        id=pre[id];
        route[cnt++]=id;
    }
    printf("%d",route[cnt-1]);
    for(int i=cnt-2;i>=0;i--){
        printf(" %d",route[i]);
    }
    printf(" %d %d
",dis[t],costs[t]);
    return 0;
}