[洛谷1342]请柬

Posted 2020-10-03 skylee的OI博客

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。

 1 #include<cstdio>
 2 #include<cctype>
 3 #include<vector>
 4 #include<ext/pb_ds/priority_queue.hpp>
 5 inline int getint() {
 6     register char ch;
 7     while(!isdigit(ch=getchar()));
 8     register int x=ch^‘0‘;
 9     while(isdigit(ch=getchar())) x=(((x<<2)+x)<<1)+(ch^‘0‘);
10     return x;
11 }
12 const long long inf=0x7fffffffffffffff;
13 const int V=1000001;
14 struct Edge {
15     int to;
16     long long w;
17 };
18 std::vector<Edge> e1[V],e2[V];
19 inline void add_edge(std::vector<Edge> *e,const int u,const int v,const long long w) {
20     e[u].push_back((Edge){v,w});
21 }
22 struct Vertex {
23     int id;
24     long long dis;
25     bool operator > (const Vertex &another) const {
26         return dis>another.dis;
27     }
28 };
29 int n;
30 long long dis1[V],dis2[V];
31 __gnu_pbds::priority_queue<Vertex,std::greater<Vertex> > q;
32 __gnu_pbds::priority_queue<Vertex,std::greater<Vertex> >::point_iterator p[V];
33 const int s=1;
34 inline void Dijkstra(long long *dis,std::vector<Edge> *e) {
35     q.clear();
36     for(register int i=1;i<=n;i++) {
37         p[i]=q.push((Vertex){i,dis[i]=(i==s)?0:inf});
38     }
39     while(q.top().dis!=inf) {
40         int x=q.top().id;
41         for(register unsigned i=0;i<e[x].size();i++) {
42             Edge &y=e[x][i];
43             if(dis[x]+y.w<dis[y.to]) {
44                 q.modify(p[y.to],(Vertex){y.to,dis[y.to]=dis[x]+y.w});
45             }
46         }
47         q.modify(p[x],(Vertex){x,inf});
48     }
49 }
50 int main() {
51     n=getint();
52     for(register int m=getint();m;m--) {
53         int u=getint(),v=getint();
54         long long w=getint();
55         add_edge(e1,u,v,w);
56         add_edge(e2,v,u,w);
57     }
58     Dijkstra(dis1,e1);
59     Dijkstra(dis2,e2);
60     long long ans=0;
61     for(register int i=2;i<=n;i++) {
62         ans+=dis1[i]+dis2[i];
63     }
64     printf("%lld\n",ans);
65     return 0;
66 }