Space Ant Time Limit: 1000MS Memory Limit: 10000K Description The most exciting space discovery occurred at the end of the 20th century. In 1999, scientists traced down an ant-like creature in the planet Y1999 and called it M11. It has only one eye o…

Technorati Tags: POJ,计算几何,凸包 初学计算几何,引入polygon后的第一个挑战--凸包 此题可用凸包算法做,只要把压入凸包的点从原集合中排除即可,最终形成图形为螺旋线. 关于凸包,具体可见凸包五法:http://blog.csdn.net/bone_ace/article/details/46239187 此处简述我O(nH)的Jarvis法(H: 凸包上点数) 过程较易于理解. //POJ1696 //凸包变形 //AC 2016.10.13 #include "cs…

Wall Time Limit: 1000MS Memory Limit: 10000K 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 would not listen to his Architect's proposals…

链接:http://poj.org/problem?id=1696 Space Ant Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 3077   Accepted: 1965 Description The most exciting space discovery occurred at the end of the 20th century. In 1999, scientists traced down an a…

Space Ant Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 2876   Accepted: 1839 Description The most exciting space discovery occurred at the end of the 20th century. In 1999, scientists traced down an ant-like creature in the planet Y19…

题目传送门 题意:一个蚂蚁一直往左边走,问最多能走多少步,且输出路径 分析:就是凸包的变形题,凸包性质,所有点都能走.从左下角开始走,不停排序.有点纠结,自己的凸包不能AC.待理解透凸包再来写.. 好像只能用卷包裹法来写,就是从一个起点出发,每次相对于起点用叉积排序,选择最外侧的点,更新起点. /************************************************ * Author :Running_Time * Created Time :2015/10/27 星期…

Space Ant Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 3316   Accepted: 2118 Description The most exciting space discovery occurred at the end of the 20th century. In 1999, scientists traced down an ant-like creature in the planet Y19…

Space Ant Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 2489   Accepted: 1567 Description The most exciting space discovery occurred at the end of the 20th century. In 1999, scientists traced down an ant-like creature in the planet Y19…

Space Ant Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 3840   Accepted: 2397 Description The most exciting space discovery occurred at the end of the 20th century. In 1999, scientists traced down an ant-like creature in the planet Y19…

Space Ant 大意:有一仅仅蚂蚁,每次都仅仅向当前方向的左边走,问蚂蚁走遍全部的点的顺序输出.開始的点是纵坐标最小的那个点,開始的方向是開始点的x轴正方向. 思路:从開始点開始,每次找剩下的点中与当前方向所形成的夹角最小的点,为下一个要走的点(好像就是犄角排序,我不是非常会),夹角就是用点积除以两个向量的距离,求一下acos值. 之前一直用叉积做,做了好久例子都没过,发现用错了... 题目挺好的,有助于理解点积与叉积 struct Point{ double x, y; int id; }…

Description The most exciting space discovery occurred at the end of the 20th century. In 1999, scientists traced down an ant-like creature in the planet Y1999 and called it M11. It has only one eye on the left side of its head and just three feet al…

Cows Time Limit: 2000MS Memory Limit: 65536K 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 forced to save money on buying fence posts b…

题意:平面上有n个点,一只蚂蚁从最左下角的点出发,只能往逆时针方向走,走过的路线不能交叉,问最多能经过多少个点. 思路:每次都尽量往最外边走,每选取一个点后对剩余的点进行极角排序.(n个点必定能走完,这是凸包的性质决定的) #include<stdio.h> #include<string.h> #include<math.h> #include<iostream> #include<algorithm> using namespace std;…

Area Time Limit: 1000MS Memory Limit: 10000K Description You are going to compute the area of a special kind of polygon. One vertex of the polygon is the origin of the orthogonal coordinate system. From this vertex, you may go step by step to the fol…

Segments Time Limit: 1000MS Memory Limit: 65536K Description Given n segments in the two dimensional space, write a program, which determines if there exists a line such that after projecting these segments on it, all projected segments have at least…

Toy Storage Time Limit: 1000MS Memory Limit: 65536K Description Mom and dad have a problem: their child, Reza, never puts his toys away when he is finished playing with them. They gave Reza a rectangular box to put his toys in. Unfortunately, Reza is…

题意: 有很多点,从最右下角的点开始走起,初始方向水平向右,然后以后每步只能向左边走,问最多能走多少个点. 解法: 贪心的搞的话,肯定每次选左边的与它夹角最小的点,然后走过去. 然后就是相当于模拟地去选点,然后计数,然后走过去.这题就这么搞定了. 我这里用了set和vector. 代码: #include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> #include &…

#include<cstdio> #include<cstring> #include<cmath> #include<iostream> #include<algorithm> #include<queue> using namespace std; ; const double PI = acos(-1.0); const double INF = 100000000.000000; struct Point{ int index…

题目大意:给出n个点的编号和坐标,按逆时针方向连接着n个点,按连接的先后顺序输出每个点的编号. 题目思路:Cross(a,b)表示a,b的叉积,若小于0:a在b的逆时针方向,若大于0a在b的顺时针方向.每次都sort一下,找出在当前点逆时针方向的最远的点.数据很小O(N*N*log(N))的复杂度0s AC了-- #include<cstdio> #include<cstdlib> #include<cmath> #include<iostream> #in…

极角排序 每次选择一个最外围的没选过的点,选择的时候需要利用极角排序进行选择 #include<cstdio> #include<cstring> #include<vector> #include<cmath> #include<queue> #include<list> #include<algorithm> using namespace std; ; struct point { int x,y; double a…

Area Time Limit: 1000MS Memory Limit: 10000K Description Being well known for its highly innovative products, Merck would definitely be a good target for industrial espionage. To protect its brand-new research and development facility the company has…

题意: 一只特别的蚂蚁,只能直走或者左转.在一个平面上,有很多株植物,这只蚂蚁每天需要进食一株,这只蚂蚁从起点为(0,miny)的点开始出发.求最多能活多少天 分析: 肯定是可以吃到所有植物的,以当前方向无限延长成直线,可以剩余的植物都在直线的左边.所以就是求上一个位置到当前位置与下一个位置与当前位置的夹角,并且使夹角最大. cos(0~pi)是单调递减的,夹角越大,cos值越小.所以我用点积来计算. #include <iostream> #include <cstdio> #i…

                                                     Space Ant 明天早上最后一科毛概了,竟然毫无复习之意,沉迷刷题无法自拔~~ 题意:说实话没有仔细看,大概是一只右眼失明(只能左转)的蚂蚁每天要吃掉一个点的食物才能活下去,现在给出n个点的坐标,求蚂蚁最多能活多少天? 一看就是循环求凸包一直往里循环,数据只有50,这让我有勇气尝试一发.样例水过了却WA了三发.讨论区说的是用点积或者叉积,但我觉得循环凸包思路也行啊.吃饭的时候突然意识到我…

凸包的算法比较形象好理解 代码写起来也比较短 所以考前看一遍应该就没什么问题了..>_< POJ1113 刚开始并没有理解为什么要用凸包,心想如果贴着城堡走不是更好吗? 突然发现题目中有要求在满足把所有点包括在内的情况下周长最短...这不就是凸包的性质吗? 而且显然如果城堡是凹的话,往里面绕一圈肯定会使周长增加... 然后可以从简单的三角形四边形推广出去,发现每个拐角-左右各90度之后所有的加和为180度 也就是在城堡周长的基础上再加一个半径为L的圆周长即是所求答案. 上次的模板写错了...应…

Art Gallery Time Limit: 1000MS Memory Limit: 10000K Description The art galleries of the new and very futuristic building of the Center for Balkan Cooperation have the form of polygons (not necessarily convex). When a big exhibition is organized, wat…

TOYS Time Limit: 2000MS Memory Limit: 65536K Description Calculate the number of toys that land in each bin of a partitioned toy box. Mom and dad have a problem - their child John never puts his toys away when he is finished playing with them. They g…

Cows Time Limit: 3000MS Memory Limit: 65536K Description Farmer John's cows have discovered that the clover growing along the ridge of the hill (which we can think of as a one-dimensional number line) in his field is particularly good. Farmer John ha…

2823: [AHOI2012]信号塔 Time Limit: 10 Sec Memory Limit: 128 MB Description 在野外训练中,为了确保每位参加集训的成员安全,实时的掌握和收集周边环境和队员信息非常重要,集训队采用的方式是在训练所在地散布N个小型传感器来收集并传递信息,这些传感器只与设在集训地中的信号塔进行通信,信号塔接收信号的覆盖范围是圆形,可以接收到所有分布在该集训区域内所有N个小型传感器(包括在该圆形的边上)发出的信号.信号塔的功率与信号塔接收范围半径的大小成…

1336: [Balkan2002]Alien最小圆覆盖 1337: 最小圆覆盖 Time Limit: 1 Sec Memory Limit: 162 MBSec Special Judge Description 给出N个点,让你画一个最小的包含所有点的圆. Input 先给出点的个数N,2<=N<=100000,再给出坐标Xi,Yi.(-10000.0<=xi,yi<=10000.0) Output 输出圆的半径,及圆心的坐标 Sample Input 6 8.0 9.0 4…

2618: [Cqoi2006]凸多边形 Time Limit: 5 Sec Memory Limit: 128 MB Description 逆时针给出n个凸多边形的顶点坐标,求它们交的面积.例如n=2时,两个凸多边形如下图: 则相交部分的面积为5.233. Input 第一行有一个整数n,表示凸多边形的个数,以下依次描述各个多边形.第i个多边形的第一行包含一个整数mi,表示多边形的边数,以下mi行每行两个整数,逆时针给出各个顶点的坐标. Output 输出文件仅包含一个实数,表示相交部分的面…