[洛谷1342]请柬

题目大意：
给定一个起点，求以其余所有点分别为必经点的最短回路之和。

思路：
建立反向图，在正反图上分别跑一遍Dijkstra，最后求和即可。
注意数据规模，要开long long不然会WA，只能拿25分。

#include<cstdio>
#include<cctype>
#include<vector>
#include<ext/pb_ds/priority_queue.hpp>
inline int getint() {
    register char ch;
    while(!isdigit(ch=getchar()));
    register int x=ch^'0';
    while(isdigit(ch=getchar())) x=(((x<<2)+x)<<1)+(ch^'0');
    return x;
}
const long long inf=0x7fffffffffffffff;
const int V=1000001;
struct Edge {
    int to;
    long long w;
};
std::vector<Edge> e1[V],e2[V];
inline void add_edge(std::vector<Edge> *e,const int u,const int v,const long long w) {
    e[u].push_back((Edge){v,w});
}
struct Vertex {
    int id;
    long long dis;
    bool operator > (const Vertex &another) const {
        return dis>another.dis;
    }
};
int n;
long long dis1[V],dis2[V];
__gnu_pbds::priority_queue<Vertex,std::greater<Vertex> > q;
__gnu_pbds::priority_queue<Vertex,std::greater<Vertex> >::point_iterator p[V];
const int s=1;
inline void Dijkstra(long long *dis,std::vector<Edge> *e) {
    q.clear();
    for(register int i=1;i<=n;i++) {
        p[i]=q.push((Vertex){i,dis[i]=(i==s)?0:inf});
    }
    while(q.top().dis!=inf) {
        int x=q.top().id;
        for(register unsigned i=0;i<e[x].size();i++) {
            Edge &y=e[x][i];
            if(dis[x]+y.w<dis[y.to]) {
                q.modify(p[y.to],(Vertex){y.to,dis[y.to]=dis[x]+y.w});
            }
        }
        q.modify(p[x],(Vertex){x,inf});
    }
}
int main() {
    n=getint();
    for(register int m=getint();m;m--) {
        int u=getint(),v=getint();
        long long w=getint();
        add_edge(e1,u,v,w);
        add_edge(e2,v,u,w);
    }
    Dijkstra(dis1,e1);
    Dijkstra(dis2,e2);
    long long ans=0;
    for(register int i=2;i<=n;i++) {
        ans+=dis1[i]+dis2[i];
    }
    printf("%lld\n",ans);
    return 0;
}