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…

/* 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…

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(…

#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…