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: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…

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: 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…

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…

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…

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) +…

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: 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…

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,…

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: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…

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…

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…

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…

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…

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>…

\(\color{#0066ff}{题目描述}\) 几千年前,有一个小王国位于太平洋的中部.王国的领土由两个分离的岛屿组成.由于洋流的冲击,两个岛屿的形状都变成了凸多边形.王国的国王想建立一座桥来连接这两个岛屿.为了把成本降到最低,国王要求你,主教,找到两个岛屿边界之间最小的距离. \(\color{#0066ff}{输入格式}\) 输入由几个测试用例组成. 每个测试用两个整数n,m(3≤n,m≤10000)开始 接下来的n行中的每一行都包含一对坐标,用来描述顶点在一个凸多边形中的位置. 下一条…

题目大意:有个国王他有一片森林,现在他想从这个森林里面砍伐一些树木做成篱笆把剩下的树木围起来,已知每个树都有不同的价值还有高度,求出来砍掉那些树可以做成篱笆把剩余的树都围起来,要使砍伐的树木的价值最小,如果有价值相同的尽量使砍伐的树木少一些. 分析:因为树木的数量是比较少的,所以枚举所有的状态,判断那个树需要砍那个树不需要,然后按照要求求出来答案即可. 代码如下: ==================================================================…

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…

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

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: 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…

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…

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…