Hdu 1007 最近点对

题目链接

Quoit Design

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 29547    Accepted Submission(s): 7741

Problem Description

Have you ever played quoit in a playground? Quoit is a game in which flat rings are pitched at some toys, with all the toys encircled awarded.
In the field of Cyberground, the position of each toy is fixed, and the ring is carefully designed so it can only encircle one toy at a time. On the other hand, to make the game look more attractive, the ring is designed to have the largest radius. Given a configuration of the field, you are supposed to find the radius of such a ring.

Assume that all the toys are points on a plane. A point is encircled by the ring if the distance between the point and the center of the ring is strictly less than the radius of the ring. If two toys are placed at the same point, the radius of the ring is considered to be 0.

 

Input

The input consists of several test cases. For each case, the first line contains an integer N (2 <= N <= 100,000), the total number of toys in the field. Then N lines follow, each contains a pair of (x, y) which are the coordinates of a toy. The input is terminated by N = 0.

 

Output

For each test case, print in one line the radius of the ring required by the Cyberground manager, accurate up to 2 decimal places. 

 

Sample Input

2 0 0 1 1 2 1 1 1 1 3 -1.5 0 0 0 0 1.5 0

 

Sample Output

0.71 0.00 0.75

 

计算最小点对的思想是：分治

注意：在下面代码中，如果使用宏定义来定义来替换取小函数min，则ＴＬＥ。

Accepted Code:

/*************************************************************************
    > File Name: 1007.cpp
    > Author: Stomach_ache
    > Mail: sudaweitong@gmail.com
    > Created Time: 2014年06月10日 星期二 19时15分21秒
    > Propose: 
 ************************************************************************/

#include <cmath>
#include <string>
#include <cstdio>
#include <vector>
#include <cstdlib>
#include <fstream>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

#define INF (1000000000.0)
#define MAX_N (100010)
//#define X first
//#define Y second
//#define min(x, y) ((x) < (y) ? (x) : (y))
//typedef pair<double, double> pii;
struct pii {
      double X, Y;
};

int n;
pii A[MAX_N], b[MAX_N];

double
min(double x, double y) {
      return x < y ? x : y;
}

bool
compare_x(pii a, pii b) {
      return a.X < b.X;
}

bool
compare_y(pii a, pii b) {
      return a.Y < b.Y;
}

double
get_dist(pii a, pii b) {
      return sqrt((a.X - b.X) * (a.X - b.X) + (a.Y - b.Y) * (a.Y - b.Y));
}

double
closest_pair(int low, int high) {
      //if (high == low) return INF;
      if (high - low == 1) return get_dist(A[low], A[high]);
      if (high - low == 2) {
          double d = get_dist(A[low], A[low+1]);
        d = min(d, get_dist(A[low], A[high]));
        d = min(d, get_dist(A[low+1], A[high]));
        return d;
    }
    int m = (low + high) / 2;
    double x = A[m].X;
    double d = min(closest_pair(low, m), closest_pair(m+1, high)); // (1)

    int len = 0;
    for (int i = low; i <= high; i++) {
          if (fabs(A[i].X - x) < d) {
            b[len++] = A[i];
        }
    }
    sort(b, b + len, compare_y);
    for (int i = 0; i < len; i++) {
          for (int j = i+1; j < len; j++) {
            double dx = b[j].X  - b[i].X; 
            double dy = b[j].Y - b[i].Y;
            if (dy >= d) break; 
            d = min(d, sqrt(dx * dx + dy * dy));
        }
    }

    return d;
}

void
solve() {
      sort(A, A + n, compare_x);
    printf("%.2f\n", closest_pair(0, n - 1) * 0.5);
}

int
main(void) {
      while (scanf("%d", &n) && n) {
          for (int i = 0; i < n; i++) 
              scanf("%lf %lf", &A[i].X, &A[i].Y);
        
        solve();
    }

    return 0;
}