平面最近点对 ZOJ 2107 POJ 3714

标签：计算几何

分治法求最近点对，模板题

第二题稍微判断一下即可

总结一下分治法基本写法：

第一部分：边界判断

第二部分：递归函数

第三部分：区间合并

第一题：

#include<stdio.h>
#include<math.h>
#include<algorithm>
using namespace std;
const double eps=1e-8;
struct Point {
	double x,y;
}p[100005],tmp[100005];
int n;
double dist(Point a,Point b){
	return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
int cmpxy(Point a,Point b){
	if(a.x!=b.x) return a.x<b.x;
	return a.y<b.y;
}
int cmpy(Point a,Point b){
	return a.y<b.y;
}
double Closest_Pair(int left,int right){
	double d=1e20;
	if(left==right) return d;
	if(left+1==right) return dist(p[left],p[right]);
	int mid=(left+right)>>1;
	double d1=Closest_Pair(left,mid);
	double d2=Closest_Pair(mid+1,right);
	d=min(d1,d2);
	int k=0;
	for(int i=left;i<=right;i++){
		if(fabs(p[mid].x-p[i].x)<d+eps) tmp[k++]=p[i];
	}
	sort(tmp,tmp+k,cmpy);
	for(int i=0;i<k;i++){
		for(int j=i+1;j<k&&tmp[j].y-tmp[i].y<d+eps;j++){
			d=min(d,dist(tmp[i],tmp[j]));
		}
	}

/*
 * Author:        lj94093
 * Created Time:  2015/5/20 星期三 下午 8:06:52
 * File Name:     Raid.cpp
 */
#include<stdio.h>
#include<string.h>
#include<vector>
#include<list>
#include<deque>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<bitset>
#include<algorithm>
#include<math.h>
using namespace std;
#define out(x) cout<<#x<<": "<<x<<endl
const double eps(1e-8);
const int maxn=100100;
const long long inf=-1u>>1;
typedef long long ll;
struct Point{
	double x,y;
	int kind;
}p[maxn*2],tmp[maxn*2];
int t,n;
int cmpxy(Point a,Point b){
	if(a.x!=b.x) return a.x<b.x;
	return a.y<b.y;
}
int cmpy(Point a,Point b){
	return a.y<b.y;
}
void init() {
	scanf("%d",&n);
	for(int i=0;i<n;i++) {
		scanf("%lf%lf",&p[i].x,&p[i].y);
		p[i].kind=0;
	}
	for(int i=0;i<n;i++) {
		scanf("%lf%lf",&p[i+n].x,&p[i+n].y);
		p[i+n].kind=1;
	}
	sort(p,p+2*n,cmpxy);
}
double dist(Point a,Point b){
	return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
double work(int left,int right) {
	double d=inf;
	if(left==right) return d;
	if(left+1==right){
		if(p[left].kind==p[right].kind) return d;
		else return dist(p[left],p[right]);
	}
	int mid=(left+right)>>1;
	d=min(work(left,mid),work(mid+1,right));
	
	int k=0;
	for(int i=left;i<=right;i++){
		if(fabs(p[i].x-p[mid].x)<d-eps) tmp[k++]=p[i];
	}
	sort(tmp,tmp+k,cmpy);
	for(int i=0;i<k;i++){
		for(int j=i+1;j<k && p[j].y-p[i].y<d-eps;j++){
			if(p[i].kind==p[j].kind) continue;
			d=min(d,dist(p[i],p[j]));
		}
	}
	
	return d;
}

int main() {
	#ifndef ONLINE_JUDGE
	freopen("in.txt","r",stdin);
	#endif
	scanf("%d",&t);
	while(t--){
		init();
		printf("%.3f\n",work(0,2*n-1)+eps);
	}
	
	return 0;
}

平面最近点对 ZOJ 2107 POJ 3714

标签：计算几何

原文地址：http://blog.csdn.net/lj94093/article/details/45876753