Constructing Roads POJ - 2421

Topic links: https://vjudge.net/problem/POJ-2421

Ideas: some villages, build some way, so that all the villages can be connected, and that all the channel length and the shortest.

The title says, some villages have been built between the road, we do not need to explain some roads built, then the pre-treatment time

Those have been built from the edge of two villages assigned to 0, then the running board when the minimum spanning tree is complete

Some road conditions have been established.

 1 #include <stdio.h>
 2 #include <iostream>
 3 #include <queue>
 4 using namespace std;
 5 
 6 const int inf = (int)1e9;
 7 const int N = 110;
 8 int g[N][N];
 9 int dis[N];
10 bool vis[N];
11 int n;
12 
13 struct node{
14     int loc;
15     int v;
16 
17     bool friend operator<(const node& a,const node& b){
18         return a.v > b.v;
19     }
20 };
21 
22 priority_queue<node > que;
23 
24 int prime(){
25 
26     que.push(node{1,0});
27     dis[1] = 0;
28 
29     while(!que.empty()){
30         int u = que.top().loc;
31         que.pop();
32 
33         vis[u] = true;
34 
35         for(int v = 1; v <= n; v++){
36             if(!vis[v] && dis[v] > g[u][v]){
37                 dis[v] = g[u][v];
38                 que.push(node{v,dis[v]});
39             }
40         }
41     }
42 
43     int ans = 0;
44     for(int i = 1; i <= n; i++){
45 
46     //    printf("%d ",dis[i]);
47         ans += dis[i];
48     }
49 
50    // printf("\n");
51 
52     return ans;
53 }
54 
55 int main(){
56 
57     scanf("%d",&n);
58 
59     for(int i = 1; i <= n; i++)
60         for(int j = 1; j <= n; j++)
61             scanf("%d",&g[i][j]);
62 
63     for(int i = 1; i <= n; i++)
64         dis[i] = inf;
65 
66     int m;
67     scanf("%d",&m);
68 
69     int u,v;
70     for(int i = 1; i <= m; i++){
71         scanf("%d%d",&u,&v);
72         G [U] [V] = G [V] [U] = 0 ; // has villages passage 
73 is      }
 74  
75      the printf ( " % D \ n- " , Prime ());
 76  
77      return  0 ;
 78 }