/* 题意: 求得n个点的凸包.然后求与凸包相距l的外圈的周长. 答案为n点的凸包周长加上半径为L的圆的周长 */ # include <stdio.h> # include <math.h> # include <string.h> # include <algorithm> using namespace std; # define PI acos(-1.0) struct node { int x; int y; }; node a[1010],res…

Wall Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3139    Accepted Submission(s): 888 Problem Description Once upon a time there was a greedy King who ordered his chief Architect to build a w…

#include<iostream> #include<iostream> #include<algorithm> #include<cmath> using namespace std; const int N=1005; const double eps=1e-8; int T,n,r,w,top; struct dian { double x,y; dian(double X=0,double Y=0) { x=X,y=Y; } dian operat…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1348 Wall Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 3386    Accepted Submission(s): 968 Problem Description Once upon a time there was a gre…

链接:http://acm.hdu.edu.cn/showproblem.php?pid=1348 Wall Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3229    Accepted Submission(s): 919 Problem Description Once upon a time there was a greedy…

传送门: POJ:点击打开链接 HDU:点击打开链接 以下是POJ上的题: Wall Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 29121   Accepted: 9746 Description Once upon a time there was a greedy King who ordered his chief Architect to build a wall around the King's cast…

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…

<题目链接> 题目大意: 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. 解题分析: 求出这些点所围成的凸包,然后所围城墙的长度就为 该凸包周长 + 以该距离为半径的圆的周长.具体证明如下: 下面的模板还没有整理好 Graham 凸包算法 #include<iostream> #include<cstdio> #include<cmath> #include&…

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…

题解:计算凸包周长 #include <iostream> #include <cmath> #include <algorithm> const int size=1000; using namespace std; struct pint{int x,y;}x[size]; int n,l,ans[size],cnt,sta[size],tail; bool cmp(pint a,pint b){return (a.y1 && !crossleft(…

http://acm.hdu.edu.cn/showproblem.php?pid=1348 造城墙问题,求出凸包加上一圈圆的周长即可 凸包模板题 #include <cstdio> #include <cstdlib> #include <cmath> #include <cstring> #include <string> #include <queue> #include <map> #include <ios…

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 help him? The diameter and length…

题意:略 思路:直接套用凸包模板 #include <iostream> #include <cstdio> #include <cmath> #include <algorithm> using namespace std; #define N 50010 struct node{ int x,y,d; }p[N]; int dist(node a,node b){ return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y…

题目链接 http://acm.hdu.edu.cn/showproblem.php?pid=1348 题意:给出一个凸包,求出与凸包距离 L的外圈周长 凸包模板题,练练Andrew算法求出凸包周长再加上半径为l的圆的周长 #include<bits/stdc++.h> #define N 1050 using namespace std; int n,l,m; ); struct P{ int x,y; bool operator < (const P &b)const{ re…

计算几何-凸包模板题目,Graham算法解. /* 1348 */ #include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> #include <cmath> #include <algorithm> using namespace std; #define MAXN 1005 typedef struct Point_t { double…

// 凸包模板 POJ1873 // n=15所以可以按位枚举求凸包,再记录数据 #include <iostream> #include <cstdio> #include <cstdlib> #include <algorithm> #include <vector> #include <math.h> using namespace std; #define LL long long typedef pair<int,in…

传送门:Surround the Trees 题意:求凸包的周长. 分析:凸包模板题,先按极角排好序后,然后根据叉积正负确定凸包. #include <stdio.h> #include <math.h> #include <algorithm> #include <string.h> #include <math.h> using namespace std; ; ; int sgn(double x) { ; ); ; } struct Po…

给出点集,和不大于L长的绳子,问能包裹住的最多点数. 考虑每个点都作为左下角的起点跑一遍极角序求凸包,求的过程中用DP记录当前以j为当前末端为结束的的最小长度,其中一维作为背包的是凸包内侧点的数量.也就是 dp[j][k]代表当前链末端为j,其内部点包括边界数量为k的最小长度.这样最后得到的一定是最优的凸包. 然后就是要注意要dp[j][k]的值不能超过L,每跑一次凸包,求个最大的点数量就好了. 和DP结合的计算几何题,主要考虑DP怎么搞 /** @Date : 2017-09-27 17:27…

Poj 2187 凸包模板求解 传送门 由于整个点数是50000,而求凸包后的点也不会很多,因此直接套凸包之后两重循环即可求解 #include <queue> #include <cmath> #include <cstdio> #include <cstring> #include <cstdlib> #include <iostream> #include <algorithm> #define ll long lo…

凸包模板--Graham扫描法 First 标签: 数学方法--计算几何 题目:洛谷P2742[模板]二维凸包/[USACO5.1]圈奶牛Fencing the Cows yyb的讲解:https://www.cnblogs.com/cjyyb/p/7260523.html 模板 #include<iostream> #include<cstdlib> #include<cstdio> #include<cmath> #include<cstring&…

http://acm.hdu.edu.cn/showproblem.php?pid=1392 题目大意: 二维平面给定n个点,用一条最短的绳子将所有的点都围在里面,求绳子的长度. 解题思路: 凸包的模板.凸包有很多的算法.这里用Adrew. 注意这几组测试数据 1 1 1 3 0 0 1 0 2 0 输出数据 0.00 2.00 #include<cmath> #include<cstdio> #include<algorithm> using namespace st…

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…

Wall Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4903    Accepted Submission(s): 1419 Problem Description Once upon a time there was a greedy King who ordered his chief Architect to build a…

Surround the Trees Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6728    Accepted Submission(s): 2556 Problem Description There are a lot of trees in an area. A peasant wants to buy a rope to…

题目:http://acm.hdu.edu.cn/showproblem.php?pid=1532 最近在学网络流,学的还不好,先不写理解了,先放模板... 我觉得写得不错的博客:http://blog.csdn.net/smartxxyx/article/details/9293665/ #include<stdio.h> #include<string.h> #include<vector> #define maxn 222 #define inf 0x3f3f3f…

题目: http://acm.hdu.edu.cn/showproblem.php?pid=2222 AC自动机模板题 我现在对AC自动机的理解还一般,就贴一下我参考学习的两篇博客的链接: http://blog.csdn.net/niushuai666/article/details/7002823 http://www.cppblog.com/menjitianya/archive/2014/07/10/207604.html #include<stdio.h> #include<s…

http://acm.hdu.edu.cn/showproblem.php?pid=2853 这道题初看了没有思路,一直想的用网络流如何解决 参考了潘大神牌题解才懂的 最大匹配问题KM 还需要一些技巧来解决最小变动, 做法是:把原先的邻接矩阵每个数扩大k倍(k>n) 为了突出原先的选择,也就是同等情况下优先选择原来的方案 给原来的方案对应矩阵内的数据+1 那么 最终得出的最大匹配值/k=真实的最大匹配 最终得出的最大匹配值%k=原来的方案采用了几个 这里的KM留下来做模板 /* 二分图最佳匹配…

1.HDU 1251 统计难题  Trie树模板题,或者map 2.总结:用C++过了,G++就爆内存.. 题意:查找给定前缀的单词数量. #include<iostream> #include<cstring> #include<cmath> #include<queue> #include<algorithm> #include<cstdio> #define max(a,b) a>b?a:b #define F(i,a,b…

此题为凸包问题模板题,题目中所给点均为整点,考虑到数据范围问题求norm()时先转换成double了,把norm()那句改成<vector>压栈即可求得凸包. 初次提交被坑得很惨,在GDB中可以完美运行A掉,到OJ上就频频RE(此处应有黑人问号) 后来发现了问题,原因是玩杂耍写了这样的代码 struct point { int x, y; point (){ scanf("%d%d", &x, &y); } ... }pt[MAXN]; 于是乎,在swap…

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…