Wall Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 28157   Accepted: 9401 Description Once upon a time there was a greedy King who ordered his chief Architect to build a wall around the King's castle. The King was so greedy, that he wo…

凸包模板--Graham扫描法 First 标签: 数学方法--计算几何 题目:洛谷P2742[模板]二维凸包/[USACO5.1]圈奶牛Fencing the Cows yyb的讲解:https://www.cnblogs.com/cjyyb/p/7260523.html 模板 #include<iostream> #include<cstdlib> #include<cstdio> #include<cmath> #include<cstring&…

There are a lot of trees in an area. A peasant wants to buy a rope to surround all these trees. So at first he must know the minimal required length of the rope. However, he does not know how to calculate it. Can you help him? The diameter and length…

题目链接:https://cn.vjudge.net/problem/POJ-3348 题意 啊模版题啊 求凸包的面积,除50即可 思路 求凸包的面积,除50即可 提交过程 AC 代码 #include <cmath> #include <cstdio> #include <vector> #include <algorithm> using namespace std; const double eps=1e-10; struct Point{ doubl…

struct P { double x, y; P(, ):x(x), y(y) {} double add(double a, double b){ ; return a+b; } P operator + (P p){ return P(add(x, p.x), add(y, p.y)); } P operator - (P p){ return P(add(x, -p.x), add(y, -p.y)); } P operator *(double d){ return P(x*d, y*…

题目链接:https://cn.vjudge.net/problem/POJ-1113 题意 给一些点,求一个能够包围所有点且每个点到边界的距离不下于L的周长最小图形的周长 思路 求得凸包的周长,再加上一个半径为L的圆的周长 提交过程 CE 注意某些OJ上cmath库里没有M_PI AC 代码 #define PI 3.1415926 #include <cmath> #include <cstdio> #include <vector> #include <al…

#include<iostream> #include<algorithm> #include<cmath> using namespace std; typedef pair<int ,int > ll; ll num,dot[1010]; int i; const double pi=3.1415926535898; ll operator -(ll a,ll b) { return make_pair(a.first-b.first,a.second-…

上模板. #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #include <cmath> #include <vector> #include <utility> #include <stack> #include <queue> #include <map> #include…

Building Fence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 171    Accepted Submission(s): 25Special Judge Problem Description Long long ago, there is a famous farmer named John. He owns a bi…

题目大意:给N个点,然后要修建一个围墙把所有的点都包裹起来,但是要求围墙距离所有的点的最小距离是L,求出来围墙的长度. 分析:如果没有最小距离这个条件那么很容易看出来是一个凸包,然后在加上一个最小距离L,那么就是在凸包外延伸长度为L,如下图,很明显可以看出来多出来的长度就是半径为L的圆的周长,所以总长度就是凸包的周长+半径为L的圆的周长. 代码如下: -------------------------------------------------------------------------…

链接: http://poj.org/problem?id=2187 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=22013#problem/E Beauty Contest Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 24254   Accepted: 7403 Description Bessie, Farmer John's prize cow, h…

A - Building Fence Time Limit:1000MS     Memory Limit:65535KB     64bit IO Format:%I64d & %I64u Submit Status Description Long long ago, there is a famous farmer named John. He owns a big farm and many cows. There are two kinds of cows on his farm, o…

Wall Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 28462   Accepted: 9498 Description Once upon a time there was a greedy King who ordered his chief Architect to build a wall around the King's castle. The King was so greedy, that he wo…

凸包算法讲解:Click Here 题目链接:https://vjudge.net/problem/POJ-1113 题意:简化下题意即求凸包的周长+2×PI×r. 思路:用graham求凸包,模板是kuangbin的,算法复杂度O(nlogn). AC code: // Author : RioTian // Time : 20/10/21 #include <algorithm> #include <cmath> #include <cstdio> #include…

题目链接 题意 : 求凸包周长+一个完整的圆周长. 因为走一圈,经过拐点时,所形成的扇形的内角和是360度,故一个完整的圆. 思路 : 求出凸包来,然后加上圆的周长 #include <stdio.h> #include <string.h> #include <iostream> #include <cmath> #include <algorithm> const double PI = acos(-1.0) ; using namespac…

Cows Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 6199   Accepted: 2822 Description Your friend to the south is interested in building fences and turning plowshares into swords. In order to help with his overseas adventure, they are f…

Beauty Contest Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 26180   Accepted: 8081 Description Bessie, Farmer John's prize cow, has just won first place in a bovine beauty contest, earning the title 'Miss Cow World'. As a result, Bess…

题目链接:https://vjudge.net/problem/POJ-1113 题意:简化下题意即求凸包的周长+2×PI×r. 思路:用graham求凸包,模板是kuangbin的. AC code: #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> using namespace std; ; const double PI=acos(-1.0); struct…

Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 25256   Accepted: 7756 Description Bessie, Farmer John's prize cow, has just won first place in a bovine beauty contest, earning the title 'Miss Cow World'. As a result, Bessie will make a…

Wall Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 31199   Accepted: 10521 Description Once upon a time there was a greedy King who ordered his chief Architect to build a wall around the King's castle. The King was so greedy, that he w…

题目链接 /* Name:nyoj-78-圈水池 Copyright: Author: Date: 2018/4/27 9:52:48 Description: Graham求凸包 zyj大佬的模板,改个输出就能用 */ #include <cstring> #include <iostream> #include <cstdio> #include <algorithm> using namespace std; struct point{ double…

Description Once upon a time there was a greedy King who ordered his chief Architect to build a wall around the King's castle. The King was so greedy, that he would not listen to his Architect's proposals to build a beautiful brick wall with a perfec…

Wall Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 26286   Accepted: 8760 Description Once upon a time there was a greedy King who ordered his chief Architect to build a wall around the King's castle. The King was so greedy, that he wo…

Description Once upon a time there was a greedy King who ordered his chief Architect to build a wall around the King's castle. The King was so greedy, that he would not listen to his Architect's proposals to build a beautiful brick wall with a perfec…

LINK 题意:给出一个简单几何,问与其边距离长为L的几何图形的周长. 思路:求一个几何图形的最小外接几何,就是求凸包,距离为L相当于再多增加上一个圆的周长(因为只有四个角).看了黑书使用graham算法极角序求凸包会有点小问题,最好用水平序比较好.或者用Melkman算法 /** @Date : 2017-07-13 14:17:05 * @FileName: POJ 1113 极角序求凸包 基础凸包.cpp * @Platform: Windows * @Author : Lweleth (…

凸包:把给定点包围在内部的.面积最小的凸多边形. Andrew算法是Graham算法的变种,速度更快稳定性也更好. 首先把全部点排序.依照第一keywordx第二keywordy从小到大排序,删除反复点后得到点序列P1...Pn. 1)把P1,P2放入凸包中,凸包中的点使用栈存储 2)从p3開始,当下一个点在凸包当前前进方向(即直线p1p2)左边的时候继续: 3)否则依次删除近期增加凸包的点,直到新点在左边. 如图,新点P18在当前前进方向P10P15的右边(使用叉积推断),因此须要从凸包上删除…

http://poj.org/problem?id=1113 Wall Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 34616   Accepted: 11821 Description Once upon a time there was a greedy King who ordered his chief Architect to build a wall around the King's castle. Th…

Wall Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 33888   Accepted: 11544 Description Once upon a time there was a greedy King who ordered his chief Architect to build a wall around the King's castle. The King was so greedy, that he w…

Wall Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2848    Accepted Submission(s): 811 Problem Description Once upon a time there was a greedy King who ordered his chief Architect to build a w…

板题hdu1348Wall 平面凸包问题是计算几何中的一个经典问题 具体就是给出平面上的多个点,求一个最小的凸多边形,使得其包含所有的点 具体形象就类似平面上有若干柱子,一个人用绳子从外围将其紧紧缠绕一圈 Graham算法 直接讲算法 我们将所有点排序,分别求出上凸壳和下凸壳,合起来就是凸包 以上凸壳为例子,我们先将最左边的点加入凸包[可以想象,最左侧的点一定在凸包上] 之后向后查找: 1.若当前凸包内只有一点,那么加入新的点 2.如果当前凸包内不止一个点,检验新加入的点与凸包最后一个点所在直线…