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…

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…

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 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…

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…

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…

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…

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;…

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…

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…

poj 1696 Space Ant 链接:http://poj.org/problem?id=1696 题意:在坐标轴上,给定n个点的 id 以及点的坐标(xi, yi),让你以最底端点开始,从右依次找出最外围的的点,形成一个螺旋状的图像.图形见题目例题,输出他们的 id. 思路:先找出最下面的点,然后依次为“据点”,对其他点进行极角排序,取出第一个点,以此点为“据点”,对其他未选择的点极角排序,选出排序后第一个点,……, 依此类推,取得 n 个点.复杂度(n^2*log(n)). 代码: #…

Space Ant Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 3967   Accepted: 2489 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 http://poj.org/problem?id=1696 Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 5371   Accepted: 3343 Description The most exciting space discovery occurred at the end of the 20th century. In 1999, scientists traced down an ant-…

Space Ant 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 all…

含极角序排序模板.   Space Ant Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 5334   Accepted: 3312 Description The most exciting space discovery occurred at the end of the 20th century. In 1999, scientists traced down an ant-like creature in th…

地址: 题目: Space Ant Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 4295   Accepted: 2697 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 pl…

/* poj 2187 Beauty Contest 凸包:寻找每两点之间距离的最大值 这个最大值一定是在凸包的边缘上的! 求凸包的算法: Andrew算法! */ #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; struct Point{ Point(){} Point(int x, int y){ this->…

/* poj3348 Cows 凸包+多边形面积 水题 floor向下取整,返回的是double */ #include<stdio.h> #include<math.h> #include <algorithm> using namespace std; const double eps = 1e-8; struct point { double x,y; }; int n; point dian[10000+10],zhan[10000+10]; /////////…

Space Ant Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 4970   Accepted: 3100 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…

POJ.2739 Sum of Consecutive Prime Numbers(水) 代码总览 #include <cstdio> #include <cstring> #include <vector> #include <cmath> #define nmax 10005 using namespace std; int prime[nmax],n; vector<int> vprime; void init() { memset(pri…

Space Ant Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 3924   Accepted: 2457 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…

POJ 2392 Space Elevator(贪心+多重背包) http://poj.org/problem?id=2392 题意: 题意:给定n种积木.每种积木都有一个高度h[i],一个数量num[i].另一个限制条件,这个积木所在的位置不能高于limit[i],问能叠起的最大高度? 分析: 本题是一道多重背包问题, 只是每一个物品的选择不只要受该种物品的数量num[i]限制, 且该物品还受到limit[i]的限制. 这里有一个贪心的结论: 我们每次背包选取物品时都应该优先放置当前limit…

@参考原文 1. 下载linux版源文件 从火狐官网下载linux版的水狐源文件压缩包,@火狐浏览器开发版(水狐)下载地址. 2. 解压下载源文件 将下载的"tar.bz2"文件解压到想要放的目录, 例如我的是/home/username/soft/firefox(很重要,注意与后面写的代码对应),解压命令为tar -jxvf xxxx.tar.bz2 (关于linux环境下的解压命令会在之后整理一份的,再附.) 3. 创建桌面启动方式 (1)编写桌面启动程序 创建办法有很多,我个人喜…