我是zz吗这么简单都写错……
一眼二分，然后判断的话是枚举点，然后计算这个点到已有联通块的最小距离，如果这个点到一些联通块的距离小于当前二分的val，则把这些联通块合并起来，这里用并查集维护，最后看这样得出的部落数是否大于k（多出来的可以直接合并）
有个非常小的优化就是不用double二分，直接把点两两之间的距离存起来排个序，直接在排序后数组里二分即可
#include<iostream>#include<cstdio>#include<cmath>#include<vector>#include<cstring>#include<algorithm>using namespace std;const int N=1005;int n,m,x[N],y[N],f[N],tot;double mn[N],d[N*N];int read(){ ???int r=0,f=1; ???char p=getchar(); ???while(p>‘9‘||p<‘0‘) ???{ ???????if(p==‘-‘) ???????????f=-1; ???????p=getchar(); ???} ???while(p>=‘0‘&&p<=‘9‘) ???{ ???????r=r*10+p-48; ???????p=getchar(); ???} ???return r*f;}double dis(double x1,double y1,double x2,double y2){ ???return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));}int zhao(int x){ ???return x==f[x]?x:f[x]=zhao(f[x]);}bool ok(double va){ ???int col=0; ???for(int i=1;i<=n;i++) ???????f[i]=i; ???for(int i=1;i<=n;i++) ???{ ???????int s=0,w; ???????for(int j=1;j<=n;j++) ???????????mn[j]=1e9; ???????for(int j=1;j<i;j++) ???????{ ???????????int nw=zhao(j); ???????????mn[nw]=min(mn[nw],dis(x[i],y[i],x[j],y[j])); ???????} ???????for(int j=1;j<i;j++) ???????????if(f[j]==j&&mn[j]<va) ???????????????s++,w=j; ???????if(s>1) ???????{ ???????????for(int j=1;j<i;j++) ???????????????if(f[j]==j&&mn[j]<va&&j!=w) ???????????????????f[j]=w; ???????????f[i]=w; ???????} ???????else if(s==1) ???????????f[i]=w; ???} ???for(int i=1;i<=n;i++) ???????if(f[i]==i) ???????????col++; ???// printf("%.2lf %d\n",va,col); ???// for(int i=1;i<=n;i++) ???????// cerr<<zhao(i)<<" "; ???// cerr<<endl; ???return col>=m;}int main(){ ???n=read(),m=read(); ???for(int i=1;i<=n;i++) ???{ ???????x[i]=read(),y[i]=read(); ???????for(int j=1;j<i;j++) ???????????d[++tot]=dis(x[i],y[i],x[j],y[j]); ???} ???sort(d+1,d+1+tot); ???int l=0,r=tot,ans; ???while(l<=r) ???{ ???????int mid=(l+r)>>1; ???????if(ok(d[mid])) ???????????l=mid+1,ans=mid; ???????else ???????????r=mid-1; ???} ???printf("%.2lf\n",d[ans]); ???return 0;}bzoj [JSOI2010]Group 部落划分 Group【二分+并查集】
原文地址：https://www.cnblogs.com/lokiii/p/9388354.html