POJ 1410 Intersection(判断线段与矩形位置关系)

系统管理员 2022-05-18 06:18 204阅读 0赞

[题目链接][Link 1]

题目大意：判断一条线段与一个实心矩形（四条边以及中间包含的部分）是否相交。

分析：线段与矩形四条边的不规范相交（交于一条线段的端点或者重合），以及线段在矩形的内部，都算作相交。这题看似简单，其实有很多比较细节的地方，比较坑，调了好久。

#include<map>
    #include<set>
    #include<cmath>
    #include<queue>
    #include<stack>
    #include<cstdio>
    #include<vector>
    #include <iomanip>
    #include<utility>
    #include<cctype>
    #include<cstring>
    #include<cstdlib>
    #include<iostream>
    #include<algorithm>
    #define inf 0x3f3f3f3f
    #define Clear(x) memset(x,0,sizeof(x))
    #define fup(i,a,b) for(int i=a;i<b;i++)
    #define rfup(i,a,b) for(int i=a;i<=b;i++)
    #define fdn(i,a,b) for(int i=a;i>b;i--)
    #define rfdn(i,a,b) for(int i=a;i>=b;i--)
    typedef long long ll;
    using namespace std;
    const double pi=acos(-1.0);
    const int maxn = 1e2+7;
    const double eps = 1e-8;
    
    struct Point{
        double x,y;
        Point(){}
        Point(double _x,double _y){x=_x;y=_y;}
    };
    
    struct Line{
        Point s,e;
        Line(){}
        Line(Point _s,Point _e){s=_s;e=_e;}
    }line[10];
    
    int sgn(double x)
    {
        if(fabs(x)<eps) return 0;
        else return x<0?-1:1;
    }
    
    double cross(Point a,Point b,Point c)
    {
        a.x-=c.x;a.y-=c.y;
        b.x-=c.x;b.y-=c.y;
        return a.x*b.y-b.x*a.y;
    }
    
    bool intersect(Line l1,Line l2){
        return(sgn(max(l1.s.x,l1.e.x)-min(l2.s.x,l2.e.x))>=0&&         //排斥实验
                sgn(max(l2.s.x,l2.e.x)-min(l1.s.x,l1.e.x))>=0&&
                sgn(max(l1.s.y,l1.e.y)-min(l2.s.y,l2.e.y))>=0&&
                sgn(max(l2.s.y,l2.e.y)-min(l1.s.y,l1.e.y))>=0&&
                sgn(cross(l2.s,l1.e,l1.s)*cross(l2.e,l1.e,l1.s)<=0)&& //跨立实验
                sgn(cross(l1.s,l2.e,l2.s)*cross(l1.e,l2.e,l2.s)<=0));
    }
    
    int main()
    {
        int T;
        scanf("%d",&T);
        double x1,y1,x2,y2,x3,y3,x4,y4;
        while(T--)
        {
            Clear(line);
            int flag=0;
    		double x1,y1,x2,y2;
    		scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3,&x4,&y4);
    		if(x3>x4)
    			swap(x3,x4);
    		if(y3<y4)
    			swap(y3,y4);
    		line[0]=Line(Point(x1,y1),Point(x2,y2));
    		line[1]=Line(Point(x3,y3),Point(x3,y4));
    		line[2]=Line(Point(x3,y4),Point(x4,y4));
    		line[3]=Line(Point(x4,y3),Point(x4,y4));
    		line[4]=Line(Point(x3,y3),Point(x4,y3));
    		for(int i=1;i<=4;i++){
                    if(intersect(line[0],line[i]))
                    {
                        flag=1;
                        break;
                    }
                }
            if(flag){}
            else if(min(x1,x2)>x3&&max(x1,x2)<x4&&min(y1,y2)>y4&&max(y1,y2)<y3)
    			flag=1;
            if(flag) printf("T\n");
            else printf("F\n");
        }
        return 0;
    }

[Link 1]: http://poj.org/problem?id=1410