标签:include hdoj bool 距离 ble col ons node 条件
题目链接:https://vjudge.net/problem/POJ-3714
题意:给定两个点集,求最短距离。
思路:在平面最近点对基础上加了个条件,我么不访用f做标记,集合1的f为1,集合2的f为-1,那么求两个点的距离时,如果a.f*b.f=-1时计算距离,否则乘积为1的话返回inf。其它就和hdoj1007一样了.
AC代码:
#include<cstdio> #include<algorithm> #include<cmath> #include<cstdlib> using namespace std; const int maxn=2e5+5; const double inf=1e30; int T,n,cnt,id[maxn]; struct node{ int x,y,f; }pt[maxn]; bool operator < (const node& a,const node& b){ if(a.x==b.x) return a.y<b.y; return a.x<b.x; } bool cmp(int a,int b){ return pt[a].y<pt[b].y; } double dist(const node& a,const node& b){ if(a.f*b.f==1) return inf; return sqrt(1.0*(a.x-b.x)*(a.x-b.x)+1.0*(a.y-b.y)*(a.y-b.y)); } double fenzhi(int l,int r){ double d=inf; if(l==r) return d; if(l+1==r) return dist(pt[l],pt[r]); int mid=(l+r)>>1; d=min(fenzhi(l,mid),fenzhi(mid+1,r)); cnt=0; int t1,t2,l1=l,r1=mid,mid1; while(l1<=r1){ mid1=(l1+r1)>>1; if(pt[mid].x-pt[mid1].x<d) r1=mid1-1; else l1=mid1+1; } t1=l1; l1=mid+1,r1=r; while(l1<=r1){ mid1=(l1+r1)>>1; if(pt[mid1].x-pt[mid].x<d) l1=mid1+1; else r1=mid1-1; } t2=r1; for(int i=t1;i<=t2;++i) id[++cnt]=i; sort(id+1,id+cnt+1,cmp); for(int i=1;i<cnt;++i) for(int j=i+1;j<=cnt&&(pt[id[j]].y-pt[id[i]].y<d);++j) d=min(d,dist(pt[id[i]],pt[id[j]])); return d; } int main(){ scanf("%d",&T); while(T--){ scanf("%d",&n); for(int i=1;i<=2*n;++i){ scanf("%d%d",&pt[i].x,&pt[i].y); if(i<=n) pt[i].f=1; else pt[i].f=-1; } sort(pt+1,pt+1+2*n); printf("%.3f\n",fenzhi(1,2*n)); } return 0; }
标签:include hdoj bool 距离 ble col ons node 条件
原文地址:https://www.cnblogs.com/FrankChen831X/p/11411688.html