Out of Hay POJ 2395 (minimum spanning tree)

The original title

Topic Link

Topic analysis

At first glance you might not see is the minimum spanning tree, but think about the subject actually required to meet the minimum spanning tree of nature, so only need to calculate the minimum spanning tree prim, record minimum spanning tree edge in the process of seeking the maximum you can.

Code

 1 #include <iostream>
 2 #include <algorithm>
 3 #include <utility>
 4 #include <cstdio>
 5 #include <cmath>
 6 #include <cstring>
 7 #include <string>
 8 #include <vector>
 9 #include <stack>
10 #include <queue>
11 #include <map>
12 #include <set>
13 
14 using namespace std;
15 typedef long long LL;
16 const int INF_INT=0x3f3f3f3f;
17 const LL INF_LL=0x3f3f3f3f3f3f3f3f;
18 
19 typedef pair<int,int> P;
20 int n,m;
21 int edge[3000][3000];
22 int used[3000];
23 int ans;
24 
25 void prim()
26 {
27     priority_queue<P,vector<P>,greater<P> > que;
28     for(int i=1;i<=n;i++)
29         if(edge[1][i]!=INF_INT) que.push(P(edge[1][i],i));
30     used[1]=true;
31     while(que.size())
32     {
33         P p=que.top();que.pop();
34         if(used[p.second]) continue;
35         used[p.second]=true;
36         ans=max(ans,p.first);
37         for(int i=1;i<=n;i++)
38         if(edge[p.second][i]!=INF_INT&&!used[i]) que.push(P(edge[p.second][i],i));
39     }
40 }
41 
42 int main()
43 {
44 //    freopen("black.in","r",stdin);
45 //    freopen("black.out","w",stdout);
46     cin>>n>>m;
47     for(int i=1;i<=n;i++)
48         for(int j=1;j<=n;j++) edge[i][j]=INF_INT;
49     for(int i=0;i<m;i++)
50     {
51         int a,b,c;
52         scanf("%d %d %d",&a,&b,&c);
53         edge[a][b]=min(edge[a][b],c);
54         edge[b][a]=min(edge[b][a],c);
55     }
56     prim();
57     printf("%d\n",ans);
58     return 0;
59 }