题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1348 求凸包周长+2*PI*L: #include <stdio.h> #include <algorithm> #include <cstring> #include <cmath> using namespace std; ; ; ); struct point { double x, y; point(){} point(double x, double…

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…

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…

1.HDU 1392 Surround the Trees 2.题意:就是求凸包周长 3.总结:第一次做计算几何,没办法,还是看了大牛的博客 #include<iostream> #include<cstring> #include<cmath> #include<queue> #include<algorithm> #include<cstdio> #include<cstdlib> #define F(i,a,b) f…

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…

Surround the Trees Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6812    Accepted Submission(s): 2594 Problem Description There are a lot of trees in an area. A peasant wants to buy a rope to…

题目链接 题意 : 让你找出最小的凸包周长 . 思路 : 用Graham求出凸包,然后对每条边求长即可. Graham详解 #include <stdio.h> #include <string.h> #include <iostream> #include <math.h> #include <algorithm> using namespace std ; struct point { int x,y ; }p[],p1[]; int n ;…

题目链接:https://vjudge.net/problem/POJ-1873 题意:n个点(2<=n<=15),给出n个点的坐标(x,y).价值v.做篱笆时的长度l,求选择哪些点来做篱笆围住另一些点,使得选出的这些点的价值和最小,如果价值和相等要求个数最小. 思路: 看来这是WF的签到题吧.数据很小,直接二进制枚举 (1<<n),然后对未选出的点求凸包的周长,仅当选出点的长度l的和>=凸包周长时才更新答案. AC code: #include<cstdio>…

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…

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: ==================================================================…

#include<iostream> #include<algorithm> #include<cmath> using namespace std; typedef pair<int ,int > ll; ll num,dot[1010]; int i; const double pi=3.1415926535898; ll operator -(ll a,ll b) { return make_pair(a.first-b.first,a.second-…

题目链接:https://vjudge.net/problem/POJ-1113 题意:简化下题意即求凸包的周长+2×PI×r. 思路:用graham求凸包,模板是kuangbin的. AC code: #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> using namespace std; ; const double PI=acos(-1.0); struct…

凸包算法讲解:Click Here 题目链接:https://vjudge.net/problem/POJ-1113 题意:简化下题意即求凸包的周长+2×PI×r. 思路:用graham求凸包,模板是kuangbin的,算法复杂度O(nlogn). AC code: // Author : RioTian // Time : 20/10/21 #include <algorithm> #include <cmath> #include <cstdio> #include…

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…

题目链接:HDU 1392 Problem Description There are a lot of trees in an area. A peasant wants to buy a rope to surround all these trees. So at first he must know the minimal required length of the rope. However, he does not know how to calculate it. Can you…

LINK 题意:类似POJ的宫殿围墙那道,只不过这道题数据稍微强了一点,有共线的情况 思路:求凸包周长加一个圆周长 /** @Date : 2017-07-20 15:46:44 * @FileName: LightOJ 1239 求凸包.cpp * @Platform: Windows * @Author : Lweleth (SoungEarlf@gmail.com) * @Link : https://github.com/ * @Version : $Id$ */ #include <s…

BZOJ_1670_[Usaco2006 Oct]Building the Moat护城河的挖掘_求凸包 Description 为了防止口渴的食蚁兽进入他的农场,Farmer John决定在他的农场周围挖一条护城河.农场里一共有N(8<=N<=5,000)股泉水,并且,护城河总是笔直地连接在河道上的相邻的两股泉水.护城河必须能保护所有的泉水,也就是说,能包围所有的泉水.泉水一定在护城河的内部,或者恰好在河道上.当然,护城河构成一个封闭的环. 挖护城河是一项昂贵的工程,于是,节约的FJ希望护城…

题目: Wall Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 119 Accepted Submission(s): 47   Problem Description Once upon a time there was a greedy King who ordered his chief Architect to build a wa…

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: 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…

A - Building Fence Time Limit:1000MS     Memory Limit:65535KB     64bit IO Format:%I64d & %I64u Submit Status Description Long long ago, there is a famous farmer named John. He owns a big farm and many cows. There are two kinds of cows on his farm, o…

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…

链接:传送门 题意:给出二维坐标轴上 n 个点,这 n 个点构成了一个城堡,国王想建一堵墙,城墙与城堡之间的距离总不小于一个数 L ,求城墙的最小长度,答案四舍五入 思路:城墙与城堡直线长度是相等的,当城堡出现拐角时,城墙必然会出现一段圆弧,这些圆弧最终会构成一个半径为 L 的圆,所以答案就是凸包的周长 + 圆的周长 balabala: 采用Jarvis步进法来求凸包,Jarvis步进法复杂度为O(nh),h为凸包顶点个数 采用Graham-Scan来求凸包,Graham - Scan 法复杂度…

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

Wall Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 33888   Accepted: 11544 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…

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…

http://poj.org/problem?id=1113 题目大意:现在要给n个点,让你修一个围墙把这些点围起来,距离最小是l 分析  :现在就是求凸包的周长然后再加上一个圆的周长 #include<stdio.h> #include<string.h> #include<stdlib.h> #include<math.h> #include<algorithm> #include<iostream> #include<qu…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1392 题意:有n棵树,每棵树有一个坐标,想用一些绳子把这些树包含起来,求需要绳子的长度: 就是求凸包的周长的,把凸包各边的长度加起来就好了:注意n<=2的情况,运用GraHam算法,时间复杂度是O(nlogn); GraHam算法的过程: #include <stdio.h> #include <algorithm> #include <cstring> #inclu…

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 /* Name:nyoj-78-圈水池 Copyright: Author: Date: 2018/4/27 9:52:48 Description: Graham求凸包 zyj大佬的模板,改个输出就能用 */ #include <cstring> #include <iostream> #include <cstdio> #include <algorithm> using namespace std; struct point{ double…