题目大意: 给定的n个点 能圈出的最大范围中 若每50平方米放一头牛 一共能放多少头 求凸包 答案就是 凸包的面积/50 向下取整 /// 求多边形面积// 凹多边形同样适用 因为点积求出的是有向面积 double areaPg() { ; ;i<k-;i++) // 以t[0]为划分顶点 res+=(t[i]-t[]).det(t[i+]-t[]); return res/2.0; } #include <cstdio> #include <algorithm> #incl…

题目大意:利用n棵树当木桩修建牛圈,知道每头牛需要50平的生存空间,求最多能放养多少头牛. 分析:赤裸裸的求凸包然后计算凸包的面积. 代码如下: ------------------------------------------------------------------------------------------------------------------------------------- #include<iostream> #include<string.h>…

/* poj 1654 Area 多边形面积 题目意思很简单,但是1000000的point开不了 */ #include<stdio.h> #include<math.h> #include<string.h> const int N=1000000+10; const double eps=1e-8; struct point { double x,y; point(){} point(double a,double b):x(a),y(b){} }; int le…

/* poj3348 Cows 凸包+多边形面积 水题 floor向下取整,返回的是double */ #include<stdio.h> #include<math.h> #include <algorithm> using namespace std; const double eps = 1e-8; struct point { double x,y; }; int n; point dian[10000+10],zhan[10000+10]; /////////…

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 forced to save money on buying fence posts by using trees as fence posts wherever possibl…

链接:http://poj.org/problem?id=3348 Cows Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 6677   Accepted: 3020 Description Your friend to the south is interested in building fences and turning plowshares into swords. In order to help with…

题目传送门 题意:告诉若干个矩形的信息,问他们在凸多边形中所占的面积比例 分析:训练指南P272,矩形面积长*宽,只要计算出所有的点,用凸包后再求多边形面积.已知矩形的中心,向量在原点参考点再旋转,角度要转换成弧度制. /************************************************ * Author :Running_Time * Created Time :2015/11/10 星期二 10:34:43 * File Name :UVA_10652.cpp…

题目传送门 题意:求凸包 + (int)求面积 / 50 /************************************************ * Author :Running_Time * Created Time :2015/11/4 星期三 11:13:29 * File Name :POJ_3348.cpp ************************************************/ #include <cstdio> #include <a…

Cows Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 7739   Accepted: 3507 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…

Cows Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 8063   Accepted: 3651 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…

#include<stdio.h> #include<algorithm> #include <cstring> using namespace std; typedef long long ll; const int MAXN = 1000008; char s[MAXN]; int dx[] = {-1, -1, -1, 0, 0, 0, 1, 1, 1}; int dy[] = {-1, 0, 1, -1, 0, 1, -1, 0, 1}; int main()…

求凸包面积.求结果后不用加绝对值,这是BBS()排序决定的. //Ps 熟练了template <class T>之后用起来真心方便= = //POJ 3348 //凸包面积 //1A 2016-10-15 #include <cstdio> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #define MAXN (10000 +…

LINK 题意:给出点集,求凸包的面积 思路:主要是求面积的考察,固定一个点顺序枚举两个点叉积求三角形面积和除2即可 /** @Date : 2017-07-19 16:07:11 * @FileName: POJ 3348 凸包面积 叉积.cpp * @Platform: Windows * @Author : Lweleth (SoungEarlf@gmail.com) * @Link : https://github.com/ * @Version : $Id$ */ #include <…

Cows Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 7038   Accepted: 3242 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…

Cows Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 8122   Accepted: 3674 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…

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…

题目链接 大意: 求凸包的面积. #include <iostream> #include <vector> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <map> #include <set> #include <string> #include <queue>…

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 forced to save money on buying fence posts by using trees as fence posts wherever possibl…

题目:http://poj.org/problem?id=1265 Sample Input 2 4 1 0 0 1 -1 0 0 -1 7 5 0 1 3 -2 2 -1 0 0 -3 -3 1 0 -3 Sample Output Scenario #1: 0 4 1.0 Scenario #2: 12 16 19.0 注意:题目给出的成对的数可不是坐标,是在x和y方向走的数量. 边界上的格点数:一条左开右闭的线段(x1, x2)->(x2, y2)上的格点数为:gcd( abs(x2-x1…

要点 凸包显然 长方形旋转较好的处理方式就是用中点的Vector加上旋转的Vector,然后每个点都扔到凸包里 多边形面积板子求凸包面积即可 #include <cstdio> #include <cmath> #include <algorithm> using namespace std; typedef double db; const db eps = 1e-9; const db PI = acos(-1.0); const int maxn = 600 *…

题目链接:POJ 1265 Problem Description Being well known for its highly innovative products, Merck would definitely be a good target for industrial espionage. To protect its brand-new research and development facility the company has installed the latest s…

题目大意:从原点开始,1-4分别代表,向右下走,向右走,向右上走,向下走,5代表回到原点,6-9代表,向上走,向左下走,向左走,向左上走.求出最后的多边形面积. 分析:这个多边形面积很明显是不规则的,可以使用向量积直接求出来面积即可. 代码如下: ------------------------------------------------------------------------------------------------------------------------------…

一个简单的用叉积求任意多边形面积的题,并不难,但我却错了很多次,double的数据应该是要转化为long long,我转成了int...这里为了节省内存尽量不开数组,直接计算,我MLE了一发...,最后看了下别人的才过,我的代码就不发了,免得误导,不得不说几何真是... 还有就是这个大神的代码,貌似G++,过不了,C++AC #include <iostream> #include <algorithm> #include <cstdio> #include <c…

Cows Time Limit: 2000MS Memory Limit: 65536K 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 forced to save money on buying fence posts b…

Area   Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 19398   Accepted: 5311 利用叉积求多边形面积,可以分解成多个三角形 利用面积公式,S=1/2*abs((x0*y1-x1*y0)+(x1*y2-x2*y1)...+(xn*yn-1-xn-1*yn)+(xn*y0-x0*yn)) ,把相邻两点和原点组成一个三角形,而总面积是这n个三角形面积的和,而三角形面积是两个相邻边向量的叉积 Descr…

原题 多组数据,到0为止. 每次给出按顺序的n个点(可能逆时针,可能顺时针),求多边形面积(保留整数) 多边形面积为依次每条边(向量)叉积/2的和 \(S=\sum _{i=1}^{n-1}p[i]*p[i+1]/2\) //逆时针为正,顺时针为负 #include<cstdio> #define N 1010 using namespace std; int n; struct hhh { double x,y; hhh() {} hhh(double _x,double _y) : x(_…

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…

题目大意: 默认从零点开始 给定n次x y上的移动距离 组成一个n边形(可能为凹多边形) 输出其 内部整点数 边上整点数 面积 皮克定理 多边形面积s = 其内部整点in + 其边上整点li / 2 - 1 那么求内部整点就是 in = s + 1 - li / 2 求整个多边形边上的整点数 //求两点ab之间的整点数 int lineSeg(P a,P b) { int dx=abs(a.x-b.x), dy=abs(a.y-b.y); && dy==) ; ; // 不包括a b两个顶…

题链: http://poj.org/problem?id=3348 题解: 计算几何,凸包,多边形面积 好吧,就是个裸题,没什么可讲的. 代码: #include<cmath> #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #define MAXN 10050 using namespace std; const double eps=1e-8…

// 此博文为迁移而来,写于2015年4月9日,不代表本人现在的观点与看法.原始地址:http://blog.sina.com.cn/s/blog_6022c4720102vxaq.html 1.前言        数学在讲解析几何,现在了解一下信息学中的计算几何也是极好的,和解析几何几乎是相通的.前面的内容比较简单,甚至自己看看就行都不值得写篇文章了,但是为了后面的内容作铺垫还是写写好了.        本章节从头到尾没有去看过别人的博文,基本上都是看的刘汝佳的<算法竞赛入门经典训练指南>,…