POJ 3348 Cows (凸包模板+凸包面积)】的更多相关文章

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…
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…
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…
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…
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: 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…
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>…