Luo Gu P1991 wireless communication network minimum spanning tree []

Topic : https://www.luogu.org/problemnew/show/P1991

The meaning of problems : the coordinate points p, s point so that there may be unrestricted communication between the s points.

To have a communication path (may be indirect), the longest section of the communication path is long twenty-two between all points.

Ideas : Recent issues are not a good look. Did not see the previous question -1, this problem did not see accurate to two decimal places, doing ah.

First built in map, and then find the minimum spanning tree in the first ps large path length.

 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 int s, n;
19 const int maxn = 505;
20 struct node{
21     int x, y;
22 }nodes[maxn];
23 double g[maxn][maxn];
24 
25 double dist(int i, int j)
26 {
27     return sqrt((nodes[i].x - nodes[j].x) * (nodes[i].x - nodes[j].x) + (nodes[i].y - nodes[j].y) * (nodes[i].y - nodes[j].y));
28 }
29 
30 bool vis[maxn];
31 double d[maxn];
32 vector<double>ans;
33 void prim(int x)
34 {
35     memset(d, 0x3f, sizeof(d));
36     d[x] = 0;
37     vis[x] = true;
38     for(int i = 1; i <= n; i++){
39         d[i] = g[x][i];
40     }
41     
42     for(int t = 1; t < n; t++){
43         double min = inf;
44         int minid;
45         for(int i = 1; i <= n; i++){
46             if(!vis[i] && d[i] < min){
47                 min = d[i];
48                 minid = i;
49             }
50         }
51         vis[minid] = true;
52         ans.push_back(min);
53         for(int i = 1; i <= n; i++){
54             if(d[i] > g[minid][i]){
55                 d[i] = g[minid][i];
56             }
57         }
58     }
59 }
60 
61 int main()
62 {
63     scanf("%d%d", &s, &n);
64     for(int i = 1; i <= n; i++){
65         scanf("%d%d", &nodes[i].x, &nodes[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     sort(ans.begin(), ans.end());
76     printf("%.2lf\n", ans[n - s - 1]);
77     return 0;
78 }