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…

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:http://acm.hdu.edu.cn/showproblem.php?pid=1392 这里介绍一种求凸包的算法:Graham.(相对于其它人的解释可能会有一些出入,但大体都属于这个算法的思想,同样可以解决凸包问题) 相对于包裹法的n*m时间,Graham算法在时间上有很大的提升,只要n*log(n)时间就够了.它的基本思想如下: 1.首先,把所有的点按照y最小优先,其次x小的优先排序 2.维护一个栈,用向量的叉积来判断新插入的点跟栈顶的点哪个在外围,如果栈顶的点在当前插入的点的…

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…

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 ;…

Surround the Trees Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 10803    Accepted Submission(s): 4187 Problem Description There are a lot of trees in an area. A peasant wants to buy a rope t…

Surround the Trees:http://acm.hdu.edu.cn/showproblem.php?pid=1392 题意: 在给定点中找到凸包,计算这个凸包的周长. 思路: 这道题找出凸包上的点后,s数组中就是按顺序的点,累加一下距离就是周长了. #include <algorithm> #include <iterator> #include <iostream> #include <cstring> #include <cstdli…

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…

Surround the Trees Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 7043    Accepted Submission(s): 2688 Problem Description There are a lot of trees in an area. A peasant wants to buy a rope to…

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 help him? The…

题面 懒得粘贴了... 大致题意:坐标系内有若干个点,问把这些点都圈起来的最小凸包周长. 题解 直接求出凸包,统计一遍答案即可 #include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<cmath> #include<queue> #include<algorithm> using namespace std; #def…

Surround the Trees Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) 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 mi…

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:给一堆二维的点,问你最少用多少距离能把这些点都围起来 思路: 凸包: 我们先找到所有点中最左下角的点p1,这个点绝对在凸包上.接下来对剩余点按照相对p1的角度升序排序,角度一样按距离升序排序.因为凸包有一个特点,从最左下逆时针走,所有线都在当前这条向量的左边,根据这个特点我们进行判断.我们从栈顶拿出两个点s[top-1],s[top],所以如果s[top-1] -> p[i] 在 s[top-1] -> s[top] 右边,那么s[top]就不是凸包上一点,就这样一直判断下去.判断左右可…

又是一道模板题 #include <iostream> #include <cstring> #include <cstdlib> #include <cstdio> #include <queue> #include <cmath> #include <algorithm> using namespace std; struct P { int x,y; double angle; } p[50010]; bool cm…

题目地址:http://acm.hdu.edu.cn/showproblem.php?pid=1392 题意:给出一些点的坐标,求最小的凸多边形把所有点包围时此多边形的周长. 解法:凸包ConvexHull 模板题 #include<iostream> #include<cstdio> #include<cstring> #include<cmath> #include<cstdlib> #include<algorithm> #de…

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): 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…

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…

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): 7203    Accepted Submission(s): 2752 Problem Description There are a lot of trees in an area. A peasant wants to buy a rope to…

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 help him? The…

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…

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…

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…

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…

Time Limit: 1000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 930    Accepted Submission(s): 200 Problem Description Can you believe it? After Gardon had solved the problem, Angel accepted him! They were sitt…

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