[洛谷1991]无线通讯网

思路：
考虑使用卫星替代最小生成树中最大的$s-1$条边，答案即为剩下的最大边。

 1 #include<cmath>
 2 #include<cstdio>
 3 #include<cctype>
 4 #include<vector>
 5 #include<algorithm>
 6 inline int getint() {
 7     char ch;
 8     while(!isdigit(ch=getchar()));
 9     int x=ch^'0';
10     while(isdigit(ch=getchar())) x=(((x<<2)+x)<<1)+(ch^'0');
11     return x;
12 }
13 inline int sqr(const int x) {
14     return x*x;
15 }
16 const int N=500;
17 struct Point {
18     int x,y;
19     friend double dist (const Point a,const Point b) {
20         return sqrt(sqr(a.x-b.x)+sqr(a.y-b.y));
21     }
22 };
23 Point p[N];
24 struct Edge {
25     int u,v;
26     double w;
27     bool operator < (const Edge &another) const {
28         return w<another.w;
29     }
30 };
31 std::vector<Edge> e;
32 class DisjointSet {
33     private:
34         int anc[N];
35         int Find(const int x) {
36             return x==anc[x]?x:anc[x]=Find(anc[x]);
37         }
38     public:
39         DisjointSet() {
40             for(int i=0;i<N;i++) anc[i]=i;
41         }
42         bool isConnected(const int x,const int y) {
43             return Find(x)==Find(y);
44         }
45         void Union(const int x,const int y) {
46             anc[Find(x)]=Find(y);
47         }
48 };
49 DisjointSet s;
50 int main() {
51     int m=getint(),n=getint();
52     for(int i=0;i<n;i++) {
53         p[i].x=getint();
54         p[i].y=getint();
55     }
56     for(int i=0;i<n;i++) {
57         for(int j=i+1;j<n;j++) {
58             e.push_back((Edge){i,j,dist(p[i],p[j])});
59         }
60     }
61     std::sort(e.begin(),e.end());
62     double ans;
63     for(unsigned i=0,cnt=n-m;i<e.size()&&cnt;i++) {
64         int u=e[i].u,v=e[i].v;
65         double w=e[i].w;
66         if(s.isConnected(u,v)) continue;
67         s.Union(u,v);
68         ans=w;
69         cnt--;
70     }
71     printf("%.2f",ans);
72     return 0;
73 }

 

转载于:https://www.cnblogs.com/skylee03/p/7399050.html