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…

题目: http://poj.org/problem?id=1113 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=22013#problem/F Wall Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 26219   Accepted: 8738 Description Once upon a time there was a greedy King who…

Wall Time Limit: 1000MS Memory Limit: 10000K 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…

[题目链接] http://poj.org/problem?id=1113 [题目大意] 给出一个城堡,要求求出距城堡距离大于L的地方建围墙将城堡围起来求所要围墙的长度 [题解] 画图易得答案为凸包的周长加一个圆的周长. [代码] #include <cstdio> #include <algorithm> #include <cmath> #include <vector> using namespace std; double EPS=1e-10; co…

http://poj.org/problem?id=1113 不多说...凸包网上解法很多,这个是用graham的极角排序,也就是算导上的那个解法 其实其他方法随便乱搞都行...我只是测一下模板... struct POINT{ double x,y; POINT(, ):x(_x),y(_y){}; }; POINT p[MAXN],s[MAXN]; double dist(POINT p1,POINT p2){ return(sqrt((p1.x-p2.x) * (p1.x-p2.x) +…

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…

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…

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…

题目链接:poj 1113   单调链凸包小结 题解:本题用到的依然是凸包来求,最短的周长,只是多加了一个圆的长度而已,套用模板,就能搞定: AC代码: #include<iostream> #include<algorithm> #include<cstdio> #include<cmath> using namespace std; int m,n; struct p { double x,y; friend int operator <(p a,…

http://poj.org/problem?id=1113 答案是凸包周长+半径为l的圆的周长... 证明?这是个坑.. #include <cstdio> #include <cstring> #include <cmath> #include <string> #include <iostream> #include <algorithm> #include <queue> #include <set>…

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 (…

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…

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…

此题为凸包问题模板题,题目中所给点均为整点,考虑到数据范围问题求norm()时先转换成double了,把norm()那句改成<vector>压栈即可求得凸包. 初次提交被坑得很惨,在GDB中可以完美运行A掉,到OJ上就频频RE(此处应有黑人问号) 后来发现了问题,原因是玩杂耍写了这样的代码 struct point { int x, y; point (){ scanf("%d%d", &x, &y); } ... }pt[MAXN]; 于是乎,在swap…

题目传送门 题意:求最短路线,使得线上任意一点离城堡至少L距离 分析:先求凸包,答案 = 凸包的长度 + 以L为半径的圆的周长 /************************************************ * Author :Running_Time * Created Time :2015/10/25 11:00:48 * File Name :POJ_1113.cpp ************************************************/ #…

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…

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: -------------------------------------------------------------------------…

题意 题目链接 给出平面上n个点的坐标.你需要建一个围墙,把所有的点围在里面,且围墙距所有点的距离不小于l.求围墙的最小长度. \(n \leqslant 10^5\) Sol 首先考虑如果没有l的限制,那么显然就是凸包的长度. 现在了距离的限制,那么显然原来建在凸包上的围墙要向外移动\(l\)的距离,同时会增加一些没有围住的位置 因为多边形的外交和为360,再根据补角的性质,画一画图就知道这一块是一个半径为\(l\)的圆. 因为总答案为凸包周长 + \(2 \pi l\) #include<c…

Wall Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 43274   Accepted: 14716 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…

题链: http://poj.org/problem?id=1113 题解: 计算几何,凸包 题意:修一圈围墙把给出的点包围起来,且被包围的点距离围墙的距离不能小于L,求围墙最短为多少. 答案其实就是等于N个点的凸包的周长+半径为L的圆的周长. 代码: #include<cmath> #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #define…

题目链接:http://poj.org/problem?id=1113 题目大意:给出点集和一个长度L,要求用最短长度的围墙把所有点集围住,并且围墙每一处距离所有点的距离最少为L,求围墙的长度. 解法:凸包+以L为半径的圆的周长.以题目中的图为例,两点之间的围墙长度之和正好就是凸包的长度,再加上每个点的拐角处(注意此处为弧,才能保证城墙距离点的距离最短)的长度. #include<iostream> #include<cstdio> #include<cstdlib>…

凸包第一题. 自己认为自己写的是Andrew 其实就是xjb写出来居然过掉了测试. 刚开始把pi定义成了int,调了半天 #include <map> #include <cmath> #include <queue> #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; #def…

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…

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…

Technorati Tags: POJ,计算几何,凸包 初学计算几何,引入polygon后的第一个挑战--凸包 此题可用凸包算法做,只要把压入凸包的点从原集合中排除即可,最终形成图形为螺旋线. 关于凸包,具体可见凸包五法:http://blog.csdn.net/bone_ace/article/details/46239187 此处简述我O(nH)的Jarvis法(H: 凸包上点数) 过程较易于理解. //POJ1696 //凸包变形 //AC 2016.10.13 #include "cs…

/* poj 2187 Beauty Contest 凸包:寻找每两点之间距离的最大值 这个最大值一定是在凸包的边缘上的! 求凸包的算法: Andrew算法! */ #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; struct Point{ Point(){} Point(int x, int y){ this->…

传送门: POJ:点击打开链接 HDU:点击打开链接 以下是POJ上的题: Wall Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 29121   Accepted: 9746 Description Once upon a time there was a greedy King who ordered his chief Architect to build a wall around the King's cast…

Wall Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 32360 Accepted: 10969 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…

题目链接:https://cn.vjudge.net/contest/276359#problem/A 题目大意:有一个国王,要在自己的城堡周围建立围墙,要求围墙能把城堡全部围起来,并且围墙距离城堡的距离至少为l,然后问你最小的消耗量. 具体思路: 将围起来城堡的围墙全部往外移,求出这些点构成的凸包,然后再加上半径为l的圆的周长,这就是最终答案. AC代码: #include<iostream> #include<stack> #include<iomanip> #in…