Geometric Shapes (poj3449多边形相交)
题意:给你一些多边形的点,判断每个多边形和那些多边形相交,编号按照字典序输出
思路:枚举每个多边形的每条边看是否相交,这里的相交是包括端点的,关键是给你正方形不相邻两个点求另外两个点怎么求,长方形给你3个点求第四个点怎么求?
因为对角线的交点为两条对角线的中点,所以
x0 + x2 = x1 + x3
y0 + y2 = y1 + y3
可以证明分割的这几个小三角形是全等的所以有
x1 - x3 = y2 - y1
y1 - y3 = x2 - x0
根据这几个式子可以推出 另外两个点的坐标
剩下的就是枚举每两个多边形的每条边是否相交
就是输入输出格式要细心点
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
struct Point
{
double x,y;
Point(double x = 0,double y = 0):x(x),y(y){}
};
typedef Point Vector;
Vector operator + (Vector a, Vector b) { return Vector(a.x+b.x,a.y+b.y) ;}
Vector operator - (Vector a, Vector b) { return Vector(a.x-b.x,a.y-b.y) ;}
Vector operator * (Vector a,double p) { return Vector(a.x*p,a.y*p) ;}
Vector operator / (Vector a,double p) { return Vector(a.x/p,a.y/p) ;}
double Dot(Vector a,Vector b) { return a.x*b.x + a.y*b.y ;}
double Length(Vector a) { return sqrt(Dot(a,a)) ;}
double Cross(Vector a, Vector b) { return a.x*b.y - a.y*b.x ;}
const double eps = 1e-8;
int dcmp(double x)
{
if(fabs(x) < eps) return 0;
else return x < 0 ? -1 : 1;
}
bool operator == (Point a,Point b)
{
return dcmp(a.x-b.x) == 0&& dcmp(a.y-b.y) == 0;
}
bool operator < (Point a,Point b)
{
return a.x < b.x || (a.x == b.x && a.y < b.y);
} bool Onsegment(Point p,Point a,Point b)
{
return dcmp(Cross(b-a,p-a)) == 0 && dcmp(Dot(b-p,a-p)) < 0 || (p == a) || (p == b);
} bool OnLine(Point p,Point a,Point b)
{
return fabs(Cross(p-a,a-b)) / Length(b-a);
} bool Segmentsection(Point a,Point b,Point c,Point d)
{
double d1 = Cross(b-a,c-a),d2 = Cross(b-a,d-a),d3 = Cross(d-c,a-c),d4 = Cross(d-c,b-c);
if(dcmp(d1)*dcmp(d2) < 0 && dcmp(d3)*dcmp(d4) < 0) return true;
else if(dcmp(d1) == 0 && Onsegment(c,a,b) ) return true;
else if(dcmp(d2) == 0 && Onsegment(d,a,b) ) return true;
else if(dcmp(d3) == 0 && Onsegment(a,c,d) ) return true;
else if(dcmp(d4) == 0 && Onsegment(b,c,d) ) return true;
else return false;
} Point Segment(Point p,Vector v,Point q,Vector w)
{
Vector u = p-q;
double t = Cross(w,u) / Cross(v,w);
return p + v*t;
} double Max(double a,double b)
{
return a > b ? a : b;
}
struct Line
{
Point s,e;
Line(Point s = 0,Point e = 0) :s(s),e(e){}
}; struct polygon
{
Point p[30];
int num;
}poly[50]; bool Ispoly(polygon a,polygon b)
{
if(a.num != 0 && b.num != 0)
{
for(int i = 0; i < a.num; i++)
{
for(int j = 0; j < b.num; j++)
{
if( Segmentsection(a.p[i],a.p[(i+1)%a.num],b.p[j],b.p[(j+1)%b.num]) )
return true;
}
}
}
return false;
}
int main()
{
char str[10],strr[20];
memset(poly,0,sizeof(poly));
while(scanf("%s",str) != EOF)
{
if(strcmp(str,".") == 0)
{
break;
}
if(strcmp(str,"-") == 0)
{
char c[30];
int k,j;
for(int i = 0; i < 26; i++)
{
k = 0;
for(j = 0; j < 26; j++)
{
if( i != j && Ispoly(poly[i],poly[j]))
{
c[k++] = j + 'A';
}
}
if(k == 0 && poly[i].num != 0)
{
printf("%c has no intersections\n",i+'A');
}
else if(poly[i].num != 0)
{
printf("%c intersects with %c",i+'A',c[0]);
if(k == 2)
{
printf(" and %c",c[1]);
}
else if(k > 2)
{
for(int m = 1; m < k-1; m++)
{
printf(", %c",c[m]);
}
printf(", and %c",c[k-1]);
}
printf("\n");
}
}
printf("\n");
memset(poly,0,sizeof(poly));
continue;
}
scanf("%s",strr);
int temp = str[0]-'A';
double x,y;
if(strcmp(strr,"square") == 0)
{
poly[temp].num = 4;
scanf(" (%lf,%lf)",&x,&y);
poly[temp].p[0].x = x, poly[temp].p[0].y = y;
scanf(" (%lf,%lf)",&x,&y);
poly[temp].p[2].x = x, poly[temp].p[2].y = y; poly[temp].p[1].x = (poly[temp].p[0].x+poly[temp].p[2].x +poly[temp].p[2].y-poly[temp].p[0].y)/2;
poly[temp].p[1].y = (poly[temp].p[0].y+poly[temp].p[2].y+poly[temp].p[0].x-poly[temp].p[2].x)/2;
poly[temp].p[3].x = (poly[temp].p[0].x+poly[temp].p[2].x +poly[temp].p[0].y-poly[temp].p[2].y)/2;
poly[temp].p[3].y = (poly[temp].p[0].y+poly[temp].p[2].y+poly[temp].p[2].x-poly[temp].p[0].x)/2; }
else if(strcmp(strr,"rectangle") == 0)
{
poly[temp].num = 4;
scanf(" (%lf,%lf)",&x,&y);
poly[temp].p[0].x = x, poly[temp].p[0].y = y;
scanf(" (%lf,%lf)",&x,&y);
poly[temp].p[1].x = x, poly[temp].p[1].y = y;
scanf(" (%lf,%lf)",&x,&y);
poly[temp].p[2].x = x, poly[temp].p[2].y = y;
poly[temp].p[3].x = (poly[temp].p[0].x + poly[temp].p[2].x - poly[temp].p[1].x);
poly[temp].p[3].y = ( poly[temp].p[2].y - poly[temp].p[1].y + poly[temp].p[0].y);
}
else if(strcmp(strr,"line") == 0)
{
poly[temp].num = 2;
scanf(" (%lf,%lf)",&x,&y);
poly[temp].p[0].x = x, poly[temp].p[0].y = y;
scanf(" (%lf,%lf)",&x,&y);
poly[temp].p[1].x = x, poly[temp].p[1].y = y;
}
else if(strcmp(strr,"polygon") == 0)
{
int n;
scanf("%d",&n);
poly[temp].num = n;
for(int i = 0; i < n; i++)
{
scanf(" (%lf,%lf)",&x,&y);
poly[temp].p[i].x = x, poly[temp].p[i].y = y;
}
}
else
{ poly[temp].num = 3;
scanf(" (%lf,%lf)",&x,&y);
poly[temp].p[0].x = x, poly[temp].p[0].y = y;
scanf(" (%lf,%lf)",&x,&y);
poly[temp].p[1].x = x, poly[temp].p[1].y = y;
scanf(" (%lf,%lf)",&x,&y);
poly[temp].p[2].x = x, poly[temp].p[2].y = y;
}
}
return 0;
}
Geometric Shapes (poj3449多边形相交)的更多相关文章
- POJ 3449 Geometric Shapes 判断多边形相交
题意不难理解,给出多个多边形,输出多边形间的相交情况(嵌套不算相交),思路也很容易想到.枚举每一个图形再枚举每一条边 恶心在输入输出,不过还好有sscanf(),不懂可以查看cplusplus网站 根 ...
- TZOJ 2560 Geometric Shapes(判断多边形是否相交)
描述 While creating a customer logo, ACM uses graphical utilities to draw a picture that can later be ...
- poj3449 Geometric Shapes【计算几何】
含[判断线段相交].[判断两点在线段两侧].[判断三点共线].[判断点在线段上]模板 Geometric Shapes Time Limit: 2000MS Memory Limit: 655 ...
- POJ 3449 Geometric Shapes(判断几个不同图形的相交,线段相交判断)
Geometric Shapes Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 1243 Accepted: 524 D ...
- POJ 3449 Geometric Shapes (求正方形的另外两点)
Geometric Shapes Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 1470 Accepted: 622 D ...
- Inheritance - SGU 129(线段与多边形相交的长度)
题目大意:给一个凸多边形(点不是按顺序给的),然后计算给出的线段在这个凸多边形里面的长度,如果在边界不计算. 分析:WA2..WA3...WA4..WA11...WA的无话可说,总之细节一定考虑清楚, ...
- hdu 5130(2014广州 圆与多边形相交模板)
题意:一个很多个点p构成的多边形,pb <= pa * k时p所占区域与多边形相交面积 设p(x,y), (x - xb)^2+(y - yb)^2 / (x - xa)^2+(y ...
- dtIntersectSegmentPoly2D 2D上的线段与多边形相交计算 产生结果:是否相交,线段跨越的开始和结束百分比,相交的边
dtIntersectSegmentPoly2D(startPos, endPos, verts, nv, tmin, tmax, segMin, segMax): http://geomalgori ...
- Geometric Shapes - POJ 3449(多边形相交)
题目大意:给一些几何图形的编号,求出来这些图形都和那些相交. 分析:输入的正方形对角线上的两个点,所以需要求出来另外两个点,公式是: x2:=(x1+x3+y3-y1)/2; y2:=(y1+y3 ...
随机推荐
- 设置tomcat启动超时,不会自动停止
tomcat启动时如果1000ms没有起来,服务就会自动停止.设置位置如下
- 【转载】一步一步搭建自己的iOS网络请求库
一步一步搭建自己的iOS网络请求库(一) 大家好,我是LastDay,很久没有写博客了,这周会分享一个的HTTP请求库的编写经验. 简单的介绍 介绍一下,NSURLSession是iOS7中新的网络接 ...
- AppStore上架规则
1. 条款和条件1.1 为App Store开发程序,开发者必须遵守 Program License Agreement (PLA).人机交互指南(HIG)以及开发者和苹果签订的任何协议和合同.以下规 ...
- APNs改动 (转)
对 APNs 的吐槽 APNs 是 Apple Push Notification service 的简称(注意 APNs 的大小写, s不需要大写). 以下是我收集的一些关于 APNs 的吐槽,你先 ...
- git 教程 git.oschina.net
官方说明:http://git.oschina.net/oschina/git-osc/wikis/%E5%B8%AE%E5%8A%A9#ssh-keys 安装完成后,在开始菜单里找到"Gi ...
- [转载]hadoop SecondNamenode详解
SecondNamenode名字看起来很象是对第二个Namenode,要么与Namenode一样同时对外提供服务,要么相当于Namenode的HA.真正的了解了SecondNamenode以后,才发现 ...
- php中的JSON中文处理
最近在PHP中要输出JSON,上网查了一下,对中文支持不太好,要不就先转成utf-8的编码,再用json_encode生成,客户端还要再utf-8转中文.对于网页已经用GB2312的服务器,不想这样折 ...
- 使用windows live writer 编辑博客日志
使用Windows Live Writer 编辑日志 一 意义 写博客日志是个需要坚持的好习惯.使用Windows Live Writer,能不受网页自带编辑器限制. Markdown支持.安装mar ...
- 将数据库字段从float修改为decimal
decimal(6,2) 可以表示0000.00~9999.99 alter table test modify aaa decimal(6,2); 则表里所有大于10000的数会被设置为9999.9 ...
- 【译】UI设计基础(UI Design Basics)--自动适配与布局(Adaptivity and Layout)(四)
2.3 自动适配与布局(Adaptivity and Layout) 2.3.1 开发成自动适配(Build In Adaptivity) 用户通常希望在自己的所有设备,各种场景中使用他们喜欢的a ...