UVA - 10245 The Closest Pair Problem

属于Divide-and-Conquer，算法课老师有讲到，就找个题目试试，思想就是不断的二分。。。考虑合并时的处理。。不解释

//============================================================================// Name        : uva10245.cpp// Author      : // Version     :// Copyright   : Your copyright notice// Description : Uva10245 in C++, Ansi-style//============================================================================#include <iostream>#include<stdlib.h>#include<stdio.h>#include<math.h>#include <limits>#include<algorithm>using namespace std;const int N=10005;struct coordination{double x,y;}a[N];int cmp(struct coordination a,struct coordination b){if(a.x==b.x)return a.y<b.y;return a.x<b.x;}int compare(const void *x,const void *y){struct coordination* a=(struct coordination* )x;struct coordination* b=(struct coordination* )y;if(a->x==b->x)return a->y-b->y;return a->x-b->x;}void print(int n){for(int i=0;i<n;i++)cout<<a[i].x<<" "<<a[i].y<<endl;}double min(double a,double b){return a>b?b:a;}double getEucleanDistance(int i,int j){return sqrt(((a[i].x-a[j].x)*(a[i].x-a[j].x)+(a[i].y-a[j].y)*(a[i].y-a[j].y)));}double closestPair(int l,int r){if(l+1==r){return (numeric_limits<double>::max)();}int mid=(l+r)/2;double dl=closestPair(l,mid);double dr=closestPair(mid,r);double d=min(dl,dr);int count;for(int i=l;i<mid;i++){if(a[i].x>=a[mid].x-d){count=0;for(int j=mid;j<r&&count<6;j++){ //比较最多不超过6个点if(a[j].x<=a[mid].x+d&&a[j].y>=a[i].y-d&&a[j].y<=a[i].y+d){count++;                    d=min(d,getEucleanDistance(i,j));}}}}return d;}int main() {int n;while(cin>>n,n){for(int i=0;i<n;i++)cin>>a[i].x>>a[i].y;sort(a,a+n,cmp);//sort t=158,while qsort t=169//qsort(a,n,sizeof(a[0]),compare);//print(n);//cout<<(numeric_limits<double>::max)()<<endl;double cd=closestPair(0,n);if(cd>=10000)cout<<"INFINITY\n";elseprintf("%.4f\n",cd);}return 0;}