zoukankan      html  css  js  c++  java
  • HDU 1007 Quoit Design

    Quoit Design

    Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 130 Accepted Submission(s): 72
     
    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
     

    题目要求求一个最近点对。

    参考了一下别人的方法。

    神分治 + 神剪支 + 神排序 就过了。

    学习了一种新的排序方法 。 有所收获~~

    #include <cstdio>
    #include <algorithm>
    #include <cstring>
    #include <cmath>
    #include <iostream>
    using namespace std;
    const int N =  100010;
    const double INF = 1e10;
    struct node{ double x, y ; }e[N];
    int n ;
    inline bool cmpx(const node &a , const node &b ){ return a.x <b.x ; }
    inline bool cmpy(const int a , const int b ){ return e[a].y < e[b].y ; }
    inline double dis( const node a , const node b ){ return sqrt ( ( a.x - b.x ) * ( a.x - b.x ) + ( a.y - b.y ) *( a.y - b.y ) ); }
    
    int a[N];
    
    double solve( int l , int r )
    {
        if( l + 1 == r ) return dis( e[l] , e[r] );
        if( l + 2 == r ) return min(  dis( e[l],e[r] ) , min( dis( e[l] ,e[l+1] ) , dis( e[l+1] , e[r]) )  );
    
        int mid = ( l + r ) >> 1;
        double ans = min( solve( l , mid ) , solve( mid + 1 , r) ) ;
        int cnt = 0 ;
        for( int i = l ; i <= r ; ++i ){
            if( e[i].x >= e[mid].x - ans && e[i].x <= e[mid].x + ans ){
                a[cnt++] = i ;
            }
        }
        sort ( a , a + cnt , cmpy );
        for( int i = 0 ; i < cnt ; ++ i ){
            for( int j = i + 1  ; j < cnt ; ++j ){
                if( e[ a[i] ].y + ans <= e[ a[j] ] .y  )break ;
                ans = min( ans , dis( e[ a[i] ] ,e[ a[j] ] ) );
            }
        }
        return ans ;
    }
    int main()
    {
        ios::sync_with_stdio(false);
    
        while( cin >> n && n ){
    
            for( int i = 0 ; i < n ; ++i ){
                cin >> e[i].x >> e[i].y ;
            }
            sort( e, e + n , cmpx );
    
            printf("%.2lf
    ", ( solve( 0 ,n-1 ) / 2.0 ) );
        }
    }
    only strive for your goal , can you make your dream come true ?
  • 相关阅读:
    HL极大算子弱(1,1)范数趋于无穷, 当维数趋于无穷
    Stein's Maximal principle
    课程: 广义相对论和波方程
    关于球乘子和BochnerRiesz乘子的相关文献
    The Hardy Uncertainty Principle
    Mar. 22 10:0011:30, 1569, "Global wellposedness for the nonlinear Schrodinger equation with derivative in energy space" by Yifei Wu
    Several questions regarding construction of functions
    通知: 强化班<调和分析与PDE>3月26日的课程 改到3月21 晚上6:009:00 地点不变
    ADO.NET Entity Framework 入门示例向导
    JavaScript 定义类方法
  • 原文地址:https://www.cnblogs.com/hlmark/p/3998952.html
Copyright © 2011-2022 走看看