P3003 [USACO10DEC] Apple Delivery

First, it is not difficult to see the shortest

Then a starting point, the two end points.

Dij running side from the starting point to compare the two from the finish, choose a small one, and then to one end as a starting point, run dij, ans plus the distance to the other end, that is the end result.

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
using namespace std;
struct node{
    int to,nxt,dis;
}e[400010];
struct edge{
    int val,nm;
    bool operator < (const edge &x) const {
        return val > x.val;
    }
};
priority_queue<edge> dij;
int head[100001],in[100001],d[100001];
int cnt;
int n,m,s1,s2,s3;
inline int read(){
    int s=0,w=1;
    char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')w=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){s=s*10+ch-'0';ch=getchar();}
    return s*w;
}
inline void add(int from,int to,int dis){
    e[++cnt]=(node){to,head[from],dis};
    head[from]=cnt;
}
inline void di(int s){
    for(int i=1;i<=n;++i)d[i]=2147483647;
    memset(in,0,sizeof(in));
    dij.push((edge){0,s});
    d[s]=0;
    while(!dij.empty()){
        int t=dij.top().nm;
        dij.pop();
        if(in[t])continue;
        in[t]=1;
        for(int i=head[t];i!=0;i=e[i].nxt){
            if(!in[e[i].to]&&d[t]+e[i].dis<d[e[i].to]){
                d[e[i].to]=d[t]+e[i].dis;
                dij.push((edge){d[e[i].to],e[i].to});
            }    
        }
    }
}
int main(){
    scanf("%d%d%d%d%d",&m,&n,&s1,&s2,&s3);
    int x,y,z;
    for(int i=1;i<=m;++i){
        scanf("%d%d%d",&x,&y,&z);
        add(x,y,z);add(y,x,z);
    }
    di(s1);
    int ans=min(d[s2],d[s3]);
    di(s2);
    ans+=d[s3];
    printf("% d " , year);
     0 ; 
}return