#堆优化dijkstra#洛谷 1828 jzoj 1287 codevs 2038 ssl 1693 香甜的黄油

版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/sugar_free_mint/article/details/82502755

题目

有$n$个点，求每个点的单源最短路径的最短和

---

分析

其实跑$n$遍$spfa$或$dijkstra$堆优化就行了

---

代码

#include <cstdio>
#include <queue>
struct node{int y,w,next;}e[2901];
int n,m,dis[801][801],ls[801],mark[801];
struct rec{
    int x,d;
    bool operator <(const rec &t)const{
        return d>t.d;
    }
};
std::priority_queue<rec>q; int ans=1000000;
int in(){
    int ans=0; char c=getchar();
    while (c<48||c>57) c=getchar();
    while (c>47&&c<58) ans=ans*10+c-48,c=getchar();
    return ans;
}
void print(int ans){
    if (ans>9) print(ans/10);
    putchar(ans%10+48);
}
int main(){
    int k=in(); n=in(); m=in();
    while (k--) mark[in()]++;
    for (register int i=1;i<=m;i++){
        register int x=in(),y=in(),w=in();
        e[i]=(node){y,w,ls[x]}; ls[x]=i;
        e[i+m]=(node){x,w,ls[y]}; ls[y]=i+m;
    }
    for (register int j=1;j<=n;j++){
        while (q.size()) q.pop();
        for (register int i=1;i<=n;i++) dis[j][i]=1000000;
        dis[j][j]=0; q.push((rec){j,0});
        while (q.size()){//堆优化
            register int x=q.top().x,d=q.top().d; q.pop();
            if (dis[j][x]!=d) continue;//懒惰删除法
            for (register int i=ls[x];i;i=e[i].next)
            if (dis[j][x]+e[i].w<dis[j][e[i].y]){//松弛操作
                dis[j][e[i].y]=dis[j][x]+e[i].w;
                q.push((rec){e[i].y,dis[j][e[i].y]});
            }
        }
    }
    for (register int i=1;i<=n;i++){
        register int sum=0;
        for (register int j=1;j<=n;j++)
            sum+=dis[i][j]*mark[j];
        ans=ans<sum?ans:sum;
    }
    return !printf("%d",ans);
}