BZOJ1821: [JSOI2010]Group 部落划分

这题乍看很吓人，其实就是一个贪心。

每次取最近的两个点所在的块合并，直到只剩下k块，输出答案。

 /**************************************************************
     Problem: 1821
     User: zhuohan123
     Language: C++
     Result: Accepted
     Time:136 ms
     Memory:18492 kb
 ****************************************************************/

 #include <iostream>
 #include <cstdio>
 #include <cmath>
 #include <algorithm>
 using namespace std;
 int f[];
 int gf(int p){return p==f[p]?p:f[p]=gf(f[p]);}
 struct point{double x,y;}p[];
 double dis(point a,point b){return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);}
 struct edge
 {
     int from,to;double c;
     friend bool operator<(edge a,edge b){return a.c<b.c;}
 }g[];int gnum;
 int main(int argc, char *argv[])
 {
     int n,k;scanf("%d%d",&n,&k);
     for(int i=;i<=n;i++)scanf("%lf%lf",&p[i].x,&p[i].y);
     for(int i=;i<=n;i++)f[i]=i;
     for(int i=;i<=n;i++)
         for(int j=;j<i;j++)
         {
             g[++gnum].from=i;
             g[gnum].to=j;
             g[gnum].c=dis(p[i],p[j]);
         }
     sort(g+,g+gnum+);
     for(int i=;i<=gnum;i++)
         if(gf(g[i].from)!=gf(g[i].to))
         {
             if(n>k){f[gf(g[i].from)]=gf(g[i].to);n--;}
             else {printf("%0.2lf\n",sqrt(g[i].c));return ;}
         }
     printf("0.00\n");
     return ;
 }