链接 枚举伞的圆心,最多只有20个,因为必须与某个现有的圆心重合. 然后再二分半径就可以了. #include <iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<stdlib.h> #include<vector> #include<cmath> #include<queue> #include<set&g…

http://acm.hdu.edu.cn/showproblem.php?pid=3264 Open-air shopping malls Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2139    Accepted Submission(s): 775 Problem Description The city of M is a…

Description The city of M is a famous shopping city and its open-air shopping malls are extremely attractive. During the tourist seasons, thousands of people crowded into these shopping malls and enjoy the vary-different shopping. Unfortunately, the…

Description The city of M is a famous shopping city and its open-air shopping malls are extremely attractive. During the tourist seasons, thousands of people crowded into these shopping malls and enjoy the vary-different shopping. Unfortunately, the…

Open-air shopping malls Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2458    Accepted Submission(s): 906 Problem Description The city of M is a famous shopping city and its open-air shopping…

题目链接: POJ:id=3831" target="_blank">http://poj.org/problem?id=3831 HDU:http://acm.hdu.edu.cn/showproblem.php?pid=3264 Description The city of M is a famous shopping city and its open-air shopping malls are extremely attractive. During the…

题目链接: http://acm.hust.edu.cn/vjudge/problem/48416 Shopping Malls Time Limit: 3000MS 问题描述 We want to create a smartphone application to help visitors of a shopping mall and you have to calculate the shortest path between pairs of locations in the mall…

题目大意是:先给你一些圆,你可以任选这些圆中的一个圆点作圆,这个圆的要求是:你画完以后.这个圆要可以覆盖之前给出的每一个圆一半以上的面积,即覆盖1/2以上每一个圆的面积. 比如例子数据,选左边还是选右边没差别,红色的圆为答案(选了左边的圆点),它覆盖了左边圆的1/2以上,也覆盖了右边圆的1/2以上. 知道了怎样求两圆面积交.那么这道题就简单了.仅仅要二分答案,然后枚举每个圆点,假设全都覆盖了1/2以上就继续二分,最后答案就得出来了. #include<iostream> #include<…

纯粹是为了改进牛吃草里的两圆交模板= =. 代码如下: #include <stdio.h> #include <algorithm> #include <string.h> #include <vector> #include <math.h> using namespace std; + ; typedef long long ll; ; const double pi = acos(-1.0); ; struct circle { doub…

传送门:http://acm.hdu.edu.cn/showproblem.php?pid=3264 题意:给你n个圆,坐标和半径,然后要在这n个圆的圆心画一个大圆,大圆与这n个圆相交的面积必须大于等于每个圆面积的一半,问你建在那个圆心半径最小,为多少. 题解:枚举这n个圆,求每个圆的最小半径,通过二分半径来求,然后取这n个的最小值即可,注意点精度就OK了. AC代码: #include <iostream> #include <cstdio> #include <cstri…