Invitation Cards HDU - 1535

原题链接

考察:最短路

思路:

        较水,正反向建图求和.

 1 #include <iostream>
 2 #include <cstring>
 3 #include <queue>
 4 using namespace std;
 5 typedef long long LL;
 6 typedef pair<LL,int> PII;
 7 const int N = 1000010;
 8 int h[N],idx,n,m,a[N],b[N],w[N];
 9 LL dist[N];
10 bool st[N];
11 struct Road{
12     int to,ne,w;
13 }road[N];
14 void add(int a,int b,int w)
15 {
16     road[idx].to = b,road[idx].w = w,road[idx].ne = h[a],h[a] = idx++;
17 }
18 void dijkstra(int s)
19 {
20     for(int i=1;i<=n+3;i++) dist[i] = 1e14,st[i] = 0;
21     dist[s] = 0;
22     priority_queue<PII,vector<PII>,greater<PII> > q;
23     q.push({0,s});
24     while(q.size())
25     {
26         PII it = q.top();
27         q.pop();
28         int u = it.second;
29         if(st[u]) continue;
30         st[u] = 1;
31         for(int i=h[u];~i;i=road[i].ne)
32         {
33             int v = road[i].to;
34             if(dist[v]>dist[u]+road[i].w)
35             {
36                 dist[v] = dist[u]+road[i].w;
37                 q.push({dist[v],v});
38             }
39         }
40     }
41 }
42 int main()
43 {
44     int T;
45     scanf("%d",&T);
46     while(T--)
47     {
48         scanf("%d%d",&n,&m);
49         memset(h,-1,sizeof h); idx = 0;
50         for(int i=1;i<=m;i++) scanf("%d%d%d",&a[i],&b[i],&w[i]),add(a[i],b[i],w[i]);
51         dijkstra(1);
52         LL ans = 0;
53         for(int i=2;i<=n;i++) ans+=dist[i];
54         memset(h,-1,sizeof h); idx = 0;
55         for(int i=1;i<=m;i++) add(b[i],a[i],w[i]);
56         dijkstra(1); idx = 0;
57         for(int i=2;i<=n;i++) ans+=dist[i];
58         printf("%lld\n",ans);
59     }
60     return 0;
61 }

 

posted @ 2021-05-01 23:40  acmloser  阅读(38)  评论(0)  编辑  收藏  举报