[原创]hdu 1007 Quoit Design [sort+分治] 寻找最小距离点对

2016-04-05 20:44:34 Tabris_ 阅读数：383

博客爬取于2020-06-14 22:44:39
以下为正文

版权声明：本文为Tabris原创文章，未经博主允许不得私自转载。
https://blog.csdn.net/qq_33184171/article/details/51068008

题目链接 http://acm.hdu.edu.cn/showproblem.php?pid=1007

Quoit Design

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 43752 Accepted Submission(s): 11379

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

题目大意 :
给你N个点 求最小距离点对

解决方案 :
因为数据过多 达到了100000 而且时间只给了1s 如果两两对比的话 一定会超时 所以想到用分治的思想解决 但第一次提交居然WA了一发 之后才想到 光分治还不行 还要对坐标排下序 我采取的是从大到小排x坐标与y坐标和 的方式 点的距离就能分布均匀

然后AC 但是跑了702ms 不开心 #_#~~

看到有人跑了0ms 差距啊 …
但是并不想在优化了
所以开始贴代码了

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# include <iostream>
# include <cstdio>
# include <cstring>
# include <algorithm>
# include <cmath>
using namespace std;
int n,m;

struct point
{
    double  x,y;
} p[100005];

double dis (const point &p1,const point &p2)
{
    return (p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y);
}

double  Minlenth(int l,int r)
{
    if(r==l) return 1000000000.0;
    if(r-l==1)
       return   dis(p[l],p[r]);
    return min(Minlenth(l,(l+r)/2),Minlenth((l+r)/2,r));
}

int cmp(point p1,point p2)
{
    return p1.x+p1.y<p2.x+p2.y;
}
int main()
{
    while(~scanf("%d",&n))
    {
        if(n==0) break;
        for(int i=0; i<n; i++)
            scanf("%lf%lf",&p[i].x,&p[i].y);
        sort(p,p+n,cmp);

        printf("%.2lf\n",sqrt(Minlenth(0,n-1))/2);
    }
    return 0;
}