标签：tchar color linker res stack queue 3.1 include make
答案是具有单调性的。 因为最近的两个部落的距离为mid，所以要是有两个野人的距离<mid，则他们一定是一个部落的。
用并查集维护各联通块，如果最后的联通块个数>=k,那么mid还可以再小点。如果<k,mid还可以再大点。
二分搞一搞就行了。
# include <cstdio> # include <cstring> # include <cstdlib> # include <iostream> # include <vector> # include <queue> # include <stack> # include <map> # include <set> # include <cmath> # include <algorithm> using namespace std; # define lowbit(x) ((x)&(-x)) # define pi 3.1415926535 # define eps 1e-9 # define MOD 45989 # define INF 1000000000 # define mem(a,b) memset(a,b,sizeof(a)) # define FOR(i,a,n) for(int i=a; i<=n; ++i) # define FO(i,a,n) for(int i=a; i<n; ++i) # define bug puts("H"); # define lch p<<1,l,mid # define rch p<<1|1,mid+1,r # define mp make_pair # define pb push_back typedef pair<int,int> PII; typedef vector<int> VI; # pragma comment(linker, "/STACK:1024000000,1024000000") typedef long long LL; int Scan() { int res=0, flag=0; char ch; if((ch=getchar())==‘-‘) flag=1; else if(ch>=‘0‘&&ch<=‘9‘) res=ch-‘0‘; while((ch=getchar())>=‘0‘&&ch<=‘9‘) res=res*10+(ch-‘0‘); return flag?-res:res; } void Out(int a) { if(a<0) {putchar(‘-‘); a=-a;} if(a>=10) Out(a/10); putchar(a%10+‘0‘); } const int N=100005; //Code begin... int a[N][2], f[N], n, k, fa[N]; int find(int x) { int s, temp; for (s=x; fa[s]>=0; s=fa[s]) ; while (s!=x) temp=fa[x], fa[x]=s, x=temp; return s; } void union_set(int x, int y) { int temp=fa[x]+fa[y]; if (fa[x]>fa[y]) fa[x]=y, fa[y]=temp; else fa[y]=x, fa[x]=temp; } bool check(double x){ int res=n, u, v; x=x*x; mem(fa,-1); FOR(i,1,n) FOR(j,i+1,n) { if ((a[i][0]-a[j][0])*(a[i][0]-a[j][0])+(a[i][1]-a[j][1])*(a[i][1]-a[j][1])>=x) continue; u=find(i), v=find(j); if (u!=v) union_set(u,v), --res; } return res>=k; } int main () { scanf("%d%d",&n,&k); FOR(i,1,n) scanf("%d%d",&a[i][0],&a[i][1]); double l=0, r=20000, mid; FOR(i,1,50) { mid=(l+r)/2; if (check(mid)) l=mid; else r=mid; } printf("%.2lf\n",mid); return 0; }
标签：tchar color linker res stack queue 3.1 include make
原文地址：http://www.cnblogs.com/lishiyao/p/6647613.html
