Circle and Points Time Limit: 5000MS   Memory Limit: 30000K Total Submissions: 8131   Accepted: 2899 Case Time Limit: 2000MS Description You are given N points in the xy-plane. You have a circle of radius one and move it on the xy-plane, so as to enc…

[题目链接] http://poj.org/problem?id=1981 [题目大意] 给出平面上一些点,问一个半径为1的圆最多可以覆盖几个点 [题解] 我们对于每个点画半径为1的圆,那么在两圆交弧上的点所画的圆,一定可以覆盖这两个点 我们对于每个点计算出其和其它点的交弧,对这些交弧计算起末位置对于圆心的极角, 对这些我们进行扫描线操作,统计最大交集数量就是答案. [代码] #include <cstdio> #include <algorithm> #include <c…

地址:http://poj.org/problem?id=1981 题目: Circle and Points Time Limit: 5000MS   Memory Limit: 30000K Total Submissions: 8198   Accepted: 2924 Case Time Limit: 2000MS Description You are given N points in the xy-plane. You have a circle of radius one and…

转载请注明出处: http://www.cnblogs.com/fraud/          ——by fraud Circle and Points Time Limit: 5000MS   Memory Limit: 30000K Total Submissions: 6850   Accepted: 2443 Case Time Limit: 2000MS Description You are given N points in the xy-plane. You have a cir…

题目链接:http://www.lydsy.com/JudgeOnline/problem.php?id=1338 1338: Pku1981 Circle and Points单位圆覆盖 Time Limit: 3 Sec  Memory Limit: 162 MBSubmit: 190  Solved: 79[Submit][Status][Discuss] Description You are given N points in the xy-plane. You have a circ…

HDU5299 圆的扫描线 && 树上删边博弈 标签(空格分隔): 未分类 给出若干个圆,可以互相嵌套但不相交或相切. 每次删去一个圆和它内部的圆,进行博弈,问谁赢. 分成两部分.首先我们要处理出给定圆的嵌套关系,然后解决博弈问题. 首先保证圆不相交切不相切的话,所有圆之间的关系形成一堆树. 那么这个问题转化为树上每次删除一个边和它的子树,删不了为输,问谁赢. 树上删边问题:每个儿子的\(sg\)为1,\(一个节点的sg=所有儿子节点的\)(sg+1)\(的异或和\). 重点是怎么处理不相…

Circle and Points Time Limit: 5000MS   Memory Limit: 30000K Total Submissions: 8346   Accepted: 2974 Case Time Limit: 2000MS Description You are given N points in the xy-plane. You have a circle of radius one and move it on the xy-plane, so as to enc…

地址:http://acm.split.hdu.edu.cn/showproblem.php?pid=3511 题目: Prison Break Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2149    Accepted Submission(s): 681 Problem Description To save Sara, Mi…

Circle and Points Time Limit: 5000MS   Memory Limit: 30000K Total Submissions: 7327   Accepted: 2651 Case Time Limit: 2000MS Description You are given N points in the xy-plane. You have a circle of radius one and move it on the xy-plane, so as to enc…

题意:给定N个点,然后给定一个半径为R的圆,问这个圆最多覆盖多少个点. 思路:在圆弧上求扫描线. 如果N比较小,不难想到N^3的算法. 一般这种覆盖问题你可以假设有两个点在圆的边界上,那么每次产生的圆去看多少个点在园内即可. 但是我们现在要更高效的做法.题目等价于,有N个半径为R的圆,问二维平面上一点最多被多少个圆覆盖.即我们可以每次求交,交的部分标记++: hihocoder1508的代码. #include<bits/stdc++.h> #define pdd pair<double…