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 <…

/* 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]; /////////…

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…

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 +…

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…

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: 9022   Accepted: 3992 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: 给几个点,用绳子圈出最大的面积养牛,输出最大面积/50 题解: Graham凸包算法的模板题 下面给出做法 1.选出x坐标最小(相同情况y最小)的点作为极点(显然他一定在凸包上) 2.其他点进行极角排序<极角指从坐标轴的某一方向逆时针旋转到向量的角度>, 极角一样按距离从近到远(可以用叉积实现) 3.用栈维护凸包上的点,将极点和极角序最小的点依次入栈 4.按顺序扫描,检查栈顶的前两个元素与这个点构成的线段是否拐向右(顺时针侧,叉积小于0) 如果满足就弹出栈顶元素,直到不满足或者栈里不足…

想当年--还是邱神给我讲的凸包来着-- #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #define eps 0.000000001 #define enter putchar('\n') #define space putchar(' ') using namespace std; typedef long long ll; template <c…

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…

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…

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>…

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…

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…

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…

Area Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 17456   Accepted: 4847 Description You are going to compute the area of a special kind of polygon. One vertex of the polygon is the origin of the orthogonal coordinate system. From thi…

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

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…

Area Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 16894   Accepted: 4698 Description You are going to compute the area of a special kind of polygon. One vertex of the polygon is the origin of the orthogonal coordinate system. From thi…

题目大意: 给你n棵树,可以用这n棵树围一个圈,然后在圈里面可以养牛,每个牛需要50平方米的空间,问最多可以养多少牛? 其实就是求一个凸包,计算凸包面积,然后除以50,然后就得到答案,直接上模板了. 凸包这一类型的题目差不多,可以作为模板使用,时间复杂度是NlogN. //Time 32ms; Memory 568K #include<iostream> #include<algorithm> using namespace std; int n; typedef struct p…

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>…

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…

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…

/* 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…

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…

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…