zoukankan      html  css  js  c++  java
  • poj 1410 线段相交判断

    http://poj.org/problem?id=1410

    Intersection
    Time Limit: 1000MS   Memory Limit: 10000K
    Total Submissions: 11329   Accepted: 2978

    Description

    You are to write a program that has to decide whether a given line segment intersects a given rectangle. 

    An example: 
    line: start point: (4,9) 
    end point: (11,2) 
    rectangle: left-top: (1,5) 
    right-bottom: (7,1) 

     
    Figure 1: Line segment does not intersect rectangle 

    The line is said to intersect the rectangle if the line and the rectangle have at least one point in common. The rectangle consists of four straight lines and the area in between. Although all input values are integer numbers, valid intersection points do not have to lay on the integer grid. 

    Input

    The input consists of n test cases. The first line of the input file contains the number n. Each following line contains one test case of the format: 
    xstart ystart xend yend xleft ytop xright ybottom 

    where (xstart, ystart) is the start and (xend, yend) the end point of the line and (xleft, ytop) the top left and (xright, ybottom) the bottom right corner of the rectangle. The eight numbers are separated by a blank. The terms top left and bottom right do not imply any ordering of coordinates.

    Output

    For each test case in the input file, the output file should contain a line consisting either of the letter "T" if the line segment intersects the rectangle or the letter "F" if the line segment does not intersect the rectangle.

    Sample Input

    1
    4 9 11 2 1 5 7 1

    Sample Output

    F

    Source

     
    ‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’‘’
    ---------------------------------------------------------------------------------------------------------------
    题目有点坑,没有说明输入的不一定是左上,以及右下,可能会是左下,以及右上
    并且,包含在矩形里也算相交,这题看题解不是好想法啊,自己想出来比较好
    #include <stdio.h>
    #include <string.h>
    #include <stdlib.h>
    #include <iostream>
    #include <algorithm>
    #include <math.h>
    
    using namespace std;
    typedef struct point
    {
        int x,y;
    }point;
    
    typedef struct line
    {
        point st,ed;
    }line;
    
    int crossProduct(point a,point b,point c)
    {
        return (c.x-a.x)*(b.y-a.y)-(c.y-a.y)*(b.x-a.x);
    }
    
    bool onSegment(point a,point b,point c)
    {
        int maxx=max(a.x,b.x);
        int maxy=max(a.y,b.y);
        int minx=min(a.x,b.x);
        int miny=min(a.y,b.y);
        if(crossProduct(a,b,c)==0&&(c.x<=maxx)&&(c.x>=minx)&&(c.y<=maxy)&&(c.y>=miny))
        {
            return true;
        }
        return false;
    }
    
    bool segIntersect(point p1,point p2,point p3,point p4)
    {
        int d1=crossProduct(p3,p4,p1);
        int d2=crossProduct(p3,p4,p2);
        int d3=crossProduct(p1,p2,p3);
        int d4=crossProduct(p1,p2,p4);
        if(d1*d2<0 && d3*d4<0)
            return true;
        if(d1==0 && onSegment(p3,p4,p1))
            return true;
        if(d2==0 && onSegment(p3,p4,p2))
            return true;
        if(d3==0 && onSegment(p1,p2,p3))
            return true;
        if(d4==0 && onSegment(p1,p2,p4))
            return true;
        return false;
    }
    
    point p[5];
    line li[5];
    int num[1005];
    
    int main()
    {
        int n,m,i,j;
        line tmp;
        int xleft,ytop,xright,ybottom;
        scanf("%d",&n);
        while(n--)
        {
            scanf("%d%d%d%d%d%d%d%d",&tmp.st.x,&tmp.st.y,&tmp.ed.x,&tmp.ed.y
                  ,&xleft,&ytop,&xright,&ybottom);
            li[0].st.x=xleft;
            li[0].st.y=ytop;
            li[0].ed.x=xleft;
            li[0].ed.y=ybottom;
    
            li[1].st.x=xleft;
            li[1].st.y=ybottom;
            li[1].ed.x=xright;
            li[1].ed.y=ybottom;
    
            li[2].st.x=xright;
            li[2].st.y=ybottom;
            li[2].ed.x=xright;
            li[2].ed.y=ytop;
    
            li[3].st.x=xright;
            li[3].st.y=ytop;
            li[3].ed.x=xleft;
            li[3].ed.y=ytop;
    
            bool flag=true;
            for(i=0;i<4;i++)
            {
                if(segIntersect(tmp.st,tmp.ed,li[i].st,li[i].ed))
                {
                    flag=false;
                    break;
                }
            }
    
            int maxx=max(xleft,xright);
            int maxy=max(ytop,ybottom);
            int minx=min(xleft,xright);
            int miny=min(ytop,ybottom);
            if(tmp.st.x<=maxx&&tmp.st.x>=minx&&tmp.st.y>=miny&&tmp.st.y<=maxy
               ||tmp.ed.x<=maxx&&tmp.ed.x>=minx&&tmp.ed.y>=miny&&tmp.ed.y<=maxy)
                flag=false;
    
            if(flag)
                printf("F
    ");
            else printf("T
    ");
        }
        return 0;
    }
    View Code
  • 相关阅读:
    DDOS攻击事件记录
    ansible批量安装zabbix客户端并实现自动发现功能
    利用api更新域名解析ip+端口转发【2】
    利用api更新域名解析ip+端口转发【1】
    网站春节开市休市设置
    获取内网路由器管理页面出口ip
    关于nginx加载配置文件的巨坑
    活动封禁刷票ip
    二十五个Python高级开发技巧,终极干货!建议收藏学习!
    一则故事带你秒懂Python GIL原理!
  • 原文地址:https://www.cnblogs.com/ccccnzb/p/3883547.html
Copyright © 2011-2022 走看看