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方法:选取一个点A,以点A为圆心做一个半径为r的圆,然后枚举另一个点B,以B为圆心做一个圆,如果这两个圆有交集,那我们在这个交集内选取一个点做半径为r的圆,这个圆就包括了A和B点,找到交集最多的区域并计算这个区域被覆盖的次数,把这个数加一就是最多能够覆盖的点个数,枚举所有的A,就可以得到最优解,剩下我想说的都在下面的图里,代码里也有相关注释; 这个题在比赛的时候我们并没有做出来,赛后看了题解才知道,由于作者的代码风格很好,所以不做修改,下面是作者的原博客地址: http://www.cnblog…
题目链接 题意 : 给你很多点和一个半径r,这个半径为r的圆能覆盖的最多的点是多少. 思路 : 对每个点做半径为 r 的圆, 求交集,交集最多的区域的被覆盖次数就是能覆盖的最多的点.贴两个链接,分析的挺好,代码写得挺好 #include <stdio.h> #include <string.h> #include <math.h> #include <iostream> #include <algorithm> using namespace s…
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爆头 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 1447    Accepted Submission(s): 601 Problem Description gameboy是一个CS高手,他最喜欢的就是扮演警察,手持M4爆土匪的头.也许这里有人没玩过CS,有必要介绍一下“爆头”这个术语:所谓爆头,就是子弹直接命中对方的头部,以秒杀…
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链接:http://acm.hdu.edu.cn/showproblem.php?pid=4631 题意:依次给你n个点,每次求出当前点中的最近点对,输出所有最近点对的和: 思路:按照x排序,然后用set维护,每次插入只更新当前点和插入点前后几个位置~ #include <iostream> #include <cstdio> #include <cmath> #include <algorithm> #include <vector> #inc…
转载请注明出处,谢谢http://blog.csdn.net/ACM_cxlove?viewmode=contents    by---cxlove 题目:给出n个城市需要去占领,有m条线段是障碍物,有p个士兵可以用.占领城市有个先后顺序,每个士兵有个背包,占领城市之后,仅能补给一次背包.问背包容量最少是多少,可以用这P个士兵完成任务,起点任意 . http://acm.hdu.edu.cn/showproblem.php?pid=4606 首先:枚举所有顶点,求一下距离,判断是否与线段相交.然…
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