uvalive 4589 Asteroids】的更多相关文章

题意:给两个凸包,凸包能旋转,求凸包重心之间的最短距离. 思路:显然两个凸包贴在一起时,距离最短.所以,先求重心,再求重心到各个面的最短距离. 三维凸包+重心求法 重心求法:在凸包内,任意枚举一点,在与凸包其他一个面组成一个三棱锥.求出每个三棱锥的重心,把三棱锥等效成一个个质点,再求整体的重心. #include<cstdio> #include<cmath> #include<cstdlib> #include<cstring> #include<q…
3862 -- Asteroids ACM-ICPC Live Archive 用给出的点求出凸包的重心,并求出重心到多边形表面的最近距离. 代码如下: #include <cstdio> #include <iostream> #include <algorithm> #include <cstring> #include <cmath> using namespace std; ; ; inline int sgn(double x) { r…
UVALive - 4108 SKYLINE Time Limit: 3000MS     64bit IO Format: %lld & %llu Submit Status uDebug Description   The skyline of Singapore as viewed from the Marina Promenade (shown on the left) is one of the iconic scenes of Singapore. Country X would a…
UVALive - 3942 Remember the Word A potentiometer, or potmeter for short, is an electronic device with a variable electric resistance. It has two terminals and some kind of control mechanism (often a dial, a wheel or a slide) with which the resistance…
UVALive - 3942 Remember the Word Neal is very curious about combinatorial problems, and now here comes a problem about words. Know- ing that Ray has a photographic memory and this may not trouble him, Neal gives it to Jiejie. Since Jiejie can’t remem…
 最小点覆盖数==最大匹配数 Asteroids Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 12678 Accepted: 6904 Description Bessie wants to navigate her spaceship through a dangerous asteroid field in the shape of an N x N grid (1 <= N <= 500). The grid co…
Asteroids Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 16211   Accepted: 8819 Description Bessie wants to navigate her spaceship through a dangerous asteroid field in the shape of an N x N grid (1 <= N <= 500). The grid contains K as…
Asteroids! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3159    Accepted Submission(s): 2106 Problem Description You're in space.You want to get home.There are asteroids.You don't want to hit…
题目传送门 /* 题意:本来有n个雕塑,等间距的分布在圆周上,现在多了m个雕塑,问一共要移动多少距离: 思维题:认为一个雕塑不动,视为坐标0,其他点向最近的点移动,四舍五入判断,比例最后乘会10000即为距离: 详细解释:http://www.cnblogs.com/zywscq/p/4268556.html */ #include <cstdio> #include <iostream> #include <algorithm> #include <cmath&…
题目链接:https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=4156 题目拷贝难度大我就不复制了. 题目大意:维护一个字符串,要求支持插入.删除操作,还有输出第 i 次操作后的某个子串.强制在线. 思路1:使用可持久化treap可破,详细可见CLJ的<可持久化数据结构的研究>. 思路2:rope大法好,详见:http…