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…

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…

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…

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…

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: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…

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…