UVA 10245 - The Closest Pair Problem

Problem J
The Closest Pair Problem
Input: standard input
Output: standard output
Time Limit: 8 seconds
Memory Limit: 32 MB

 

Given a set of points in a two dimensional space, you will have to find the distance between the closest two points.

 

Input

 

The input file contains several sets of input. Each set of input starts with an integer N (0<=N<=10000), which denotes the number of points in this set. The next N line contains the coordinates of N two-dimensional points. The first of the two numbers denotes the X-coordinate and the latter denotes the Y-coordinate. The input is terminated by a set whose N=0. This set should not be processed. The value of the coordinates will be less than 40000 and non-negative.

 

Output

 

For each set of input produce a single line of output containing a floating point number (with four digits after the decimal point) which denotes the distance between the closest two points. If there is no such two points in the input whose distance is less than 10000, print the line INFINITY.

 

Sample Input

3
0 0
10000 10000
20000 20000
5
0 2
6 67
43 71
39 107
189 140
0

 

Sample Output

INFINITY
36.2215

  很经典的算法，翻开任何一本算法书籍，分治算法那一节 。  

  STL的接口都是左闭又开 。  [)  。

  合并的地方用到了  原地归并排序（上知网找） 。

 

 

#include <iostream>
#include <string>
#include <string.h>
#include <map>
#include <stdio.h>
#include <algorithm>
#include <queue>
#include <vector>
#include <math.h>
#include <set>
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
using namespace std ;
typedef long long LL ;
const int Max_N = 10008 ;
const double INF = 100000.0 ;
typedef pair<double ,double> P ;
bool cmp_y(P A ,P B){
     return A.second < B.second ;
}
int N ;
P A[Max_N] ;
double close_pair(P *a ,int n){
    if(n == 1)
      return INF ;
    int m = n>>1 ;
    int x = a[m].first ;
    double dist = min(close_pair(a,m),close_pair(a+m,n-m)) ;
    inplace_merge(a,a+m,a+n,cmp_y) ;   // 原地归并排序
    vector<P>vec ;
    for(int i = 0 ; i < n ; i++){
        if(fabs(a[i].first - x)>=dist)
            continue ;
        for(int j = vec.size() - 1 ; j >= 0 ; j--){
            double dx = a[i].first - vec[j].first ;
            double dy = a[i].second - vec[j].second ;
            if(dy >= dist)
               break ;
            dist = min(dist,sqrt(dx*dx + dy*dy)) ;
        }
        vec.push_back(a[i]) ;
    }
    return dist ;
}
int main(){
   int T ;
   while(scanf("%d",&N)&&N){
        for(int i = 0 ; i < N ; i++)
            scanf("%lf%lf",&A[i].first,&A[i].second) ;
        sort(A,A+N) ;
        double dist = close_pair(A,N) ;
        if(dist - 10000 <= 1e-9)
           printf("%.4lf\n",close_pair(A , N)) ;
        else
           puts("INFINITY") ;
   }
   return 0 ;
}