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 ************************************************/ #…

题目传送门 题意:一个蚂蚁一直往左边走,问最多能走多少步,且输出路径 分析:就是凸包的变形题,凸包性质,所有点都能走.从左下角开始走,不停排序.有点纠结,自己的凸包不能AC.待理解透凸包再来写.. 好像只能用卷包裹法来写,就是从一个起点出发,每次相对于起点用叉积排序,选择最外侧的点,更新起点. /************************************************ * Author :Running_Time * Created Time :2015/10/27 星期…

题目传送门 题意:求两点的距离平方的最大值 分析:凸包模板题 /************************************************ * Author :Running_Time * Created Time :2015/10/25 9:31:11 * File Name :A.cpp ************************************************/ #include <cstdio> #include <algorithm&…

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 (…

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,…

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 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 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) +…