Title: https://www.luogu.org/problemnew/show/P1265

There are coordinates of n cities. Each city applies for and builds its own nearest city road. After that, they are regarded as a city.

Continue until all cities are one. Ask how many roads to build at least. In the title, if a ring is formed, the shortest one should be removed.

But cities ask for it one by one, and then it becomes a city. How can it form a ring

Train of thought: the meaning of this question is very strange. I didn't understand for a long time at the beginning, but I thought that it was required to generate the largest trees. In fact, it's just a minimal spanning tree.

Because there are many points, the graph will not be saved. When prim is used directly, dist function is used.

Note that in the dist function, x and y are integers. First, convert them to double. Some of the questions don't matter. This one is WA. Pay attention to the following.

 1 #include<cstdio>
 2 #include<cstdlib>
 3 #include<map>
 4 #include<set>
 5 #include<cstring>
 6 #include<algorithm>
 7 #include<vector>
 8 #include<cmath> 
 9 #include<stack>
10 #include<queue>
11 #include<iostream>
12 
13 #define inf 0x3f3f3f3f
14 using namespace std;
15 typedef long long LL;
16 typedef pair<int, int> pr;
17 
18 const int maxn = 5005;
19 int n;
20 struct node{
21     int x, y;
22 }city[maxn];
23 //double g[maxn][maxn];
24 double d[maxn];
25 bool vis[maxn];
26 double ans = 0;
27 
28 double dist(int i, int j)
29 {
30     return sqrt((double)(city[i].x - city[j].x) * (city[i].x - city[j].x) + (double)(city[i].y - city[j].y) * (city[i].y - city[j].y));
31 }
32 
33 void prim(int st)
34 {
35     vis[st] = true;
36     d[st] = 0;
37     for(int i = 1; i <= n; i++){
38         d[i] = dist(st, i);
39     }
40     for(int t = 1; t < n; t++){
41         double min = inf;
42         int minid;
43         for(int i = 1; i <= n; i++){
44             if(!vis[i] && d[i] < min){
45                 min = d[i];
46                 minid = i;
47             }
48         }
49         vis[minid] = true;
50         ans += min;
51         for(int i = 1; i <= n; i++){
52             if(d[i] > dist(minid, i)){
53                 d[i] = dist(minid, i);
54             }
55         }
56     }
57 }
58 
59 int main()
60 {
61     scanf("%d", &n);
62     memset(d, 0x3f, sizeof(d)); 
63     for(int i = 1; i <= n; i++){
64         //par[i] = i;
65         scanf("%d%d", &city[i].x, &city[i].y);
66     }
67 //    for(int i = 1; i <= n; i++){
68 //        for(int j = i + 1; j <= n; j++){
69 //            g[i][j] = g[j][i] = dist(i, j);
70 //            //printf("%lf\n", g[i][j]); 
71 //        }
72 //    }
73     
74     prim(1);
75     printf("%.2lf\n", ans);
76     return 0;
77 }