poj 3335（半平面交）
链接：http://poj.org/problem?id=3335 //大牛们常说的测模板题
----------------------------------------------------------------
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 5158 | Accepted: 2061 |
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 edges of the polygon. We want to place a rotating scoreboard somewhere in the hall such that a spectator sitting anywhere on the boundary of the hall can view the scoreboard (i.e., his line of sight is not blocked by a wall). Note that if the line of sight of a spectator is tangent to the polygon boundary (either in a vertex or in an edge), he can still view the scoreboard. You may view spectator's seats as points along the boundary of the simple polygon, and consider the scoreboard as a point as well. Your program is given the corners of the hall (the vertices of the polygon), and must check if there is a location for the scoreboard (a point inside the polygon) such that the scoreboard can be viewed from any point on the edges of the polygon.
Input
The first number in the input line, T is the number of test cases. Each test case is specified on a single line of input in the form n x1 y1 x2 y2 ... xn yn where n (3 ≤ n ≤ 100) is the number of vertices in the polygon, and the pair of integers xi yi sequence specify the vertices of the polygon sorted in order.
Output
The output contains T lines, each corresponding to an input test case in that order. The output line contains either YES or NO depending on whether the scoreboard can be placed inside the hall conforming to the problem conditions.
Sample Input
2
4 0 0 0 1 1 1 1 0
8 0 0 0 2 1 2 1 1 2 1 2 2 3 2 3 0
Sample Output
YES
NO
Source
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
#include <math.h> using namespace std; #define eps 1e-8
#define MAXX 105
typedef struct
{
double x;
double y;
}point; point p[MAXX],s[MAXX]; bool dy(double x,double y) {return x>y+eps; }
bool xy(double x,double y) {return x<y-eps; }
bool dyd(double x,double y){return x>y-eps; }
bool xyd(double x,double y){return x<y+eps; }
bool dd(double x,double y) {return fabs(x-y)<eps; } double crossProduct(point a,point b,point c)
{
return (c.x-a.x)*(b.y-a.y)-(c.y-a.y)*(b.x-a.x);
} point IntersectPoint(point u1,point u2,point v1,point v2)
{
point ans=u1;
double t=((u1.x-v1.x)*(v1.y-v2.y)-(u1.y-v1.y)*(v1.x-v2.x))/
((u1.x-u2.x)*(v1.y-v2.y)-(u1.y-u2.y)*(v1.x-v2.x));
ans.x += (u2.x-u1.x)*t;
ans.y += (u2.y-u1.y)*t;
return ans;
} void cut(point p[],point s[],int n,int &len)
{
point tp[MAXX];
p[n]=p[];
for(int i=; i<=n; i++)
{
tp[i]=p[i];
}
int cp=n,tc;
for(int i=; i<n; i++)
{
tc=;
for(int k=; k<cp; k++)
{
if(dyd(crossProduct(p[i],p[i+],tp[k]),0.0))
s[tc++]=tp[k];
if(xy(crossProduct(p[i],p[i+],tp[k])*
crossProduct(p[i],p[i+],tp[k+]),0.0))
s[tc++]=IntersectPoint(p[i],p[i+],tp[k],tp[k+]);
}
s[tc]=s[];
for(int k=; k<=tc; k++)
tp[k]=s[k];
cp=tc;
}
len=cp;
} int main()
{
int n,m,i,j;
scanf("%d",&n);
while(n--)
{
scanf("%d",&m);
for(i=; i<m; i++)
scanf("%lf%lf",&p[i].x,&p[i].y);
int len;
cut(p,s,m,len);
if(len)printf("YES\n");
else printf("NO\n");
}
return ;
}
poj 3335（半平面交）的更多相关文章
- poj 1755 半平面交+不等式
Triathlon Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 6461 Accepted: 1643 Descrip ...
- poj 1279 半平面交核面积
Art Gallery Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 6668 Accepted: 2725 Descr ...
- poj 3335 /poj 3130/ poj 1474 半平面交 判断核是否存在 / poj1279 半平面交 求核的面积
/*************** poj 3335 点序顺时针 ***************/ #include <iostream> #include <cmath> #i ...
- poj 3525 半平面交求多边形内切圆最大半径【半平面交】+【二分】
<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, ...
- POJ 3525 /// 半平面交 模板
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 ...
- POJ 3525 半平面交+二分
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> # ...
- POJ 3335 Rotating Scoreboard 半平面交求核
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: ...
- poj 3335 Rotating Scoreboard（半平面交）
Rotating Scoreboard Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 6420 Accepted: 25 ...
- 三道半平面交测模板题 Poj1474 Poj 3335 Poj 3130
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 <cst ...
随机推荐
- window 10 用composer安装 基于 Laravel 5.3 的 Coaster CMS Curl错误77解决
安装教程来自 http://laravelacademy.org/post/6547.html 1.系统环境windows 10 nodejs3.9.2 composer1.2.4(国内镜像) Cmd ...
- RHEL6 某业务用户ulimit -a命令找不到
最终确定是shell环境问题,临时改为/bin/bash即可查看. 1.问题现象 # su - jingyu $ id uid=(jingyu) gid=(jingyu) (jingyu) $ uli ...
- editPlus修改默认的文件编码
- Rabin-Miller算法
首先附上matrix67大神的讲解: --------------------------------------------------------------------------------- ...
- 使用kd-tree加速k-means
0.目录 前置知识 思路介绍 详述 1 确定h的中心点 2 算法步骤 java实现 1.前置知识 本文内容基于<Accelerating exact k-means algorithms wit ...
- BZOJ 1191: [HNOI2006]超级英雄Hero(二分图匹配)
云神说他二分图匹配从来都是用网络流水过去的...我要发扬他的精神.. 这道题明显是二分图匹配.网络流的话可以二分答案+最大流.虽然跑得很慢.... -------------------------- ...
- Js-Html 前端系列--点击非Div区域隐藏Div
最近做项目要用到三字码下拉,调用一个插件,但是滚动条不能点击. 后来看另一个插件后突发奇想,在点击其他区域的时候隐藏这个Div就可以了. 背景:输入文字,弹出下拉Div,点击或者拉动Div滚动条选择, ...
- 系统启动 之 Linux系统启动概述(1)
随着智能终端功能的越来越庞大,与之,硬件配置越来越高,开机时间却越来越长.人们在享受强大功能的同时,对冗长的智能终端的开机时间却越来越缺乏耐心. 为了"取悦"用户,需要提供较好的用 ...
- 【BZOJ2434】阿狸的打字机（AC自动机，树状数组）
[BZOJ2434]阿狸的打字机(AC自动机,树状数组) 先写个暴力: 每次打印出字符串后,就插入到\(Trie\)树中 搞完后直接搭\(AC\)自动机 看一看匹配是怎么样的: 每次沿着\(AC\)自 ...
- 我们为什么要搞长沙.NET技术社区（三）
我们为什么要搞长沙.NET技术社区(三) 万事先从饭局开始是中华民族留下来的一个优秀的传统美德. 昨天晚餐时间,长沙 .net 技术社区的主要发起人员进行了一番小聚,同时也作为一个非正式会议,对社区发 ...