Triathlon Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 6461   Accepted: 1643 Description Triathlon is an athletic contest consisting of three consecutive sections that should be completed as fast as possible as a whole. The first sect…

Art Gallery Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 6668   Accepted: 2725 Description The art galleries of the new and very futuristic building of the Center for Balkan Cooperation have the form of polygons (not necessarily conve…

<题目链接> 题目大意:给出一个四面环海的凸多边形岛屿,求出这个岛屿中的点到海的最远距离. 解题分析: 仔细思考就会发现,其实题目其实就是让我们求该凸多边形内内切圆的最大半径是多少.但是,这个最大半径,没有什么比较好的求法,于是,我们可以想到二分答案求半径.对于二分的半径,我们可以将该凸多边形的边界向内平移 r 的距离,然后再用半平面交法,用这些平移后的直线去切割原凸多边形,如果最终切得的区域不为空,则二分枚举更大的半径,反之减小枚举的半径.知道恰好围成的区域为空(或恰好不为空)为止. #in…

题目大意: 给定n,接下来n行逆时针给定小岛的n个顶点 输出岛内离海最远的点与海的距离 半平面交模板题 将整个小岛视为由许多半平面围成 那么以相同的比例缩小这些半平面 一直到缩小到一个点时 那个点就是离海最远的点 #include <cstdio> #include <cmath> #include <vector> #include <algorithm> using namespace std; ; double add(double a,double…

/*************** poj 3335 点序顺时针 ***************/ #include <iostream> #include <cmath> #include <algorithm> using namespace std; ; const double maxn = 0x7f7f7f7f; int dcmp(double x){ if(fabs(x)<eps) ; else ?-:; } struct point { double…

二分所能形成圆的最大距离,然后将每一条边都向内推进这个距离,最后所有边组合在一起判断时候存在内部点 #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #include <cmath> using namespace std; #define N 105 #define ll long long #define eps 1e-7 int dcm…

LINK 题意:给出一个多边形,求是否存在核. 思路:比较裸的题,要注意的是求系数和交点时的x和y坐标不要搞混...判断核的顶点数是否大于1就行了 /** @Date : 2017-07-20 19:55:49 * @FileName: POJ 3335 半平面交求核.cpp * @Platform: Windows * @Author : Lweleth (SoungEarlf@gmail.com) * @Link : https://github.com/ * @Version : $Id$…

Triathlon Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 6912   Accepted: 1790 Description Triathlon is an athletic contest consisting of three consecutive sections that should be completed as fast as possible as a whole. The first sect…

Description Triathlon is an athletic contest consisting of three consecutive sections that should be completed as fast as possible as a whole. The first section is swimming, the second section is riding bicycle and the third one is running. The speed…

Triathlon Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 4733   Accepted: 1166 Description Triathlon is an athletic contest consisting of three consecutive sections that should be completed as fast as possible as a whole. The first sect…

链接:http://poj.org/problem?id=3335     //大牛们常说的测模板题 ---------------------------------------------------------------- Rotating Scoreboard Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 5158   Accepted: 2061 Description This year, ACM/ICPC…

第一道半平面交,只会写N^2. 将每条边化作一个不等式,ax+by+c>0,所以要固定顺序,方便求解. 半平面交其实就是对一系列的不等式组进行求解可行解. 如果某点在直线右侧,说明那个点在区域内,否则出现在左边,就可能会有交点,将交点求出加入. //#pragma comment(linker, "/STACK:16777216") //for c++ Compiler #include <stdio.h> #include <iostream> #inc…

Rotating Scoreboard Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 6420   Accepted: 2550 Description This year, ACM/ICPC World finals will be held in a hall in form of a simple polygon. The coaches and spectators are seated along the ed…

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…

求半平面交的算法是zzy大神的排序增量法. ///Poj 1474 #include <cmath> #include <algorithm> #include <cstdio> using namespace std; ; //点 class Point { public: double x, y; Point(){} Point(double x, double y):x(x),y(y){} bool operator < (const Point &…

http://poj.org/problem?id=1474 解法同POJ 1279 A一送一 缺点是还是O(n^2) ...nlogn的过几天补上... /********************* Template ************************/ #include <set> #include <map> #include <list> #include <cmath> #include <ctime> #include…

http://poj.org/problem?id=1279 顺时针给你一个多边形...求能看到所有点的面积...用半平面对所有边取交即可,模版题 这里的半平面交是O(n^2)的算法...比较逗比...暴力对每条线段做半平面交...要注意的地方写在注释里了...顺序写反了卡了我好久 /********************* Template ************************/ #include <set> #include <map> #include <…

题目链接:POJ 3130 Problem Description After counting so many stars in the sky in his childhood, Isaac, now an astronomer and a mathematician uses a big astronomical telescope and lets his image processing program count stars. The hardest part of the prog…

题目链接:POJ 2451 Problem Description Prince Remmarguts solved the CHESS puzzle successfully. As an award, Uyuw planned to hold a concert in a huge piazza named after its great designer Ihsnayish. The piazza in UDF - United Delta of Freedom's downtown wa…

uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1058 半平面交求面积最值.直接枚举C(20,8)的所有情况即可. 代码如下: #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #include <…

按逆时针顺序给出n个点,求它们组成的多边形的最大内切圆半径. 二分这个半径,将所有直线向多边形中心平移r距离,如果半平面交不存在那么r大了,否则r小了. 平移直线就是对于向量ab,因为是逆时针的,向中心平移就是向向量左手边平移,求出长度为r方向指向向量左手边的向量p,a+p指向b+p就是平移后的向量. 半平面交就是对于每个半平面ax+by+c>0,将当前数组里的点(一开始是所有点)带入,如果满足条件,那么保留该点,否则,先看i-1号点是否满足条件,如果满足,那么将i-1和i点所在直线和直线ax+…

题目链接 题意 : 给你一个多边形,问你里边能够盛的下的最大的圆的半径是多少. 思路 :先二分半径r,半平面交向内推进r.模板题 #include <stdio.h> #include <string.h> #include <iostream> #include <math.h> ; using namespace std ; struct node { double x; double y ; } p[],temp[],newp[];//p是最开始的多边…

题目链接 题意 : 求一个多边形的核的面积. 思路 : 半平面交求多边形的核,然后在求面积即可. #include <stdio.h> #include <string.h> #include <iostream> #include <math.h> using namespace std ; struct node { double x; double y ; } p[],temp[],newp[];//p是最开始的多边形的每个点,temp是中间过程中临时…

题目链接 题意 : 给你一个多边形,问你在多边形内部是否存在这样的点,使得这个点能够看到任何在多边形边界上的点. 思路 : 半平面交求多边形内核. 半平面交资料 关于求多边形内核的算法 什么是多边形的内核? 它是平面简单多边形的核是该多边形内部的一个点集,该点集中任意一点与多边形边界上一点的连线都处于这个多边形内部.就是一个在一个房子里面放一个摄像 头,能将所有的地方监视到的放摄像头的地点的集合即为多边形的核. 如上图,第一个图是有内核的,比如那个黑点,而第二个图就不存在内核了,无论点在哪里,总…

给出三个半平面交的裸题. 不会的上百度上谷(gu)歌(gou)一下. 毕竟学长的语文是体育老师教的.(卡格玩笑,别当真.) 这种东西明白就好,代码可以当模板. //poj1474 Video Surveillance //点集默认顺时针 //算法参考:http://www.cnblogs.com/huangxf/p/4067763.html #include<cstdio> #include<cmath> using namespace std; ; struct point{ d…

Most Distant Point from the Sea Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 5153   Accepted: 2326   Special Judge Description The main land of Japan called Honshu is an island surrounded by the sea. In such an island, it is natural t…

<题目链接> 题目大意: 给出一个凸多边形的房间,根据风水要求,把两个圆形地毯铺在房间里,不能折叠,不能切割,可以重叠.问最多能覆盖多大空间,输出两个地毯的圆心坐标.多组解输出其中一个,题目保证至少可以放入一个圆. 解题分析: 因为放置的圆不能超出多边形的边界,所以先将该凸多边形的各个边长向内平移 r 的距离,然后对这些平移后的直线用半平面交去切割原多边形,切割后得到的区域就是两圆圆心所在的区域,然后遍历这个切割后的多边形的各个顶点,距离最远的两个顶点就是这两圆的圆心. #include<…

Feng Shui Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 3743   Accepted: 1150   Special Judge Description Feng shui is the ancient Chinese practice of placement and arrangement of space to achieve harmony with the environment. George h…

Most Distant Point from the Sea Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 3476   Accepted: 1596   Special Judge Description The main land of Japan called Honshu is an island surrounded by the sea. In such an island, it is natural t…

Description The children's game Hotter Colder is played as follows. Player A leaves the room while player B hides an object somewhere in the room. Player A re-enters at position (0,0) and then visits various other positions about the room. When playe…