HDU1007(最近点对)

Quoit Design

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 54667    Accepted Submission(s): 14401

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

 

Author

CHEN, Yue

 

Source

ZJCPC2004

 

//2017-08-09
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>
#define mid ((l+r)>>1)

using namespace std;

const int N = 100010;
struct Point{
    double x, y;
}P[N], p1[N], p2[N];
int n;

bool cmp_x(Point a, Point b){
    return a.x < b.x;
}

bool cmp_y(Point a, Point b){
    return a.y < b.y;
}

double distance(Point *a, Point *b){
    return sqrt((a->x - b->x)*(a->x - b->x) + (a->y - b->y)*(a->y - b->y));
}

//分治，solve(l, r)表示区间[l, r]内的最近点对，solve(l, r) = min(solve(l, mid), solve(mid+1, r), 跨左右子区间的最近点对)
double solve(int l, int r){
    if(l >= r)return 0;
    if(r - l == 1)return distance(&P[l], &P[r]);
    if(r - l == 2)return min(distance(&P[l], &P[l+1]), distance(&P[l+1], &P[r]));
    double ans = min(solve(l, mid), solve(mid+1, r));
    //暴力x坐标与mid的x坐标相差不超过当前最优解ans的点
    int m = 0;
    for(int i = l; i <= r; i++){
        if(fabs(P[mid].x - P[i].x) <= ans){
            p1[m++] = P[i];
        }
    }
    sort(p1, p1+m, cmp_y);
    for(int i = 0; i < m; i++){
        for(int j = i+1; j < m; j++){
            if(p1[j].y - p1[i].y > ans)break;
            ans = min(ans, distance(&p1[i], &p1[j]));
        }
    }
    return ans;
}

int main()
{
    //freopen("dataIn.txt", "r", stdin);
    while(scanf("%d", &n)!=EOF && n){
        for(int i = 0; i < n; i++)
              scanf("%lf%lf", &P[i].x, &P[i].y);
        sort(P, P+n, cmp_x);
        printf("%.2lf
", solve(0, n-1)/2);
    }

    return 0;
}