POJ 1410 Intersection 数据错误

题目要求判断一条线段和一个矩形是否相交，或者是否在矩形里面（题目好像没说？）

思路就是直接暴力判断和矩形四条边是否相交，和线段的坐标是否在矩形的坐标范围即可。

然后题目的数据，(xleft,ytop) 和 (xright,ybottom)不是按顺序给出的，需要自己判断下顺序。

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;

#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
const int maxn = 5000+20;
struct coor
{
    double x,y;
    coor(){}
    coor(double xx,double yy):x(xx),y(yy){}
    double operator ^(coor rhs) const //计算叉积（向量积）
    {
        return x*rhs.y - y*rhs.x;
    }
    coor operator -(coor rhs) const //坐标相减，a-b得到向量ba
    {
        return coor(x-rhs.x,y-rhs.y);
    }
    double operator *(coor rhs) const //数量积
    {
        return x*rhs.x + y*rhs.y;
    }
}a,b;
struct Line
{
    coor point1,point2;
    Line(){}
    Line(coor xx,coor yy):point1(xx),point2(yy){}
}line[maxn],seg;
const double eps = 1e-14;
bool same (double a,double b)
{
    return fabs(a-b)<eps;
}
bool OnSegment (coor a,coor b,coor c) //判断点C是否在线段ab上
{
    double min_x = min(a.x,b.x), min_y = min(a.y,b.y);
    double max_x = max(a.x,b.x), max_y = max(a.y,b.y);
    if (c.x>=min_x && c.x<=max_x && c.y>=min_y && c.y<=max_y) return true;
    else return false;
}
bool SegmentIntersect (coor a,coor b,coor c,coor d)
{
    double d1 = (b-a)^(d-a); //direction(a,b,d);以a为起点，计算ab和ab的叉积
    double d2 = (b-a)^(c-a);
    double d3 = (d-c)^(a-c);
    double d4 = (d-c)^(b-c);
    if (d1*d2<0 && d3*d4<0) return true;
    else if (same(d1,0) && OnSegment(a,b,d)) return true;
    else if (same(d2,0) && OnSegment(a,b,c)) return true;
    else if (same(d3,0) && OnSegment(c,d,a)) return true;
    else if (same(d4,0) && OnSegment(c,d,b)) return true;
    else return false;
}
void work ()
{
    scanf("%lf%lf%lf%lf",&seg.point1.x,&seg.point1.y,&seg.point2.x,&seg.point2.y);
    scanf("%lf%lf%lf%lf",&a.x,&a.y,&b.x,&b.y);
    if (a.x > b.x) swap(a.x,b.x);
    if (a.y < b.y) swap(a.y,b.y);
    line[1] = Line(a,coor(b.x,a.y));
    line[2] = Line(coor(a.x,b.y),b);
    line[3] = Line(a,coor(a.x,b.y));
    line[4] = Line(coor(b.x,a.y),b);
    for (int i=1;i<=4;++i)
    {
        if (SegmentIntersect(seg.point1,seg.point2,line[i].point1,line[i].point2))
        {
            printf ("T\n");
            return ;
        }
    }
    if (seg.point1.x>=a.x&&seg.point1.x<=b.x&&seg.point2.x>=a.x&&seg.point2.x<=b.x
      &&seg.point1.y>=b.y&&seg.point1.y<=a.y&&seg.point2.y>=b.y&&seg.point2.y<=a.y)
      {
          printf ("T\n");
          return ;
      }
    printf ("F\n");
    return ;
}

int main()
{
#ifdef local
    freopen("data.txt","r",stdin);
#endif
    int t;
    cin>>t;
    while(t--) work();
    return 0;
}

 

 

然而这题应该用判断点在多边形里比较好

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;

#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
const int maxn = 5000+20;
struct coor
{
    double x,y;
    coor(){}
    coor(double xx,double yy):x(xx),y(yy){}
    double operator ^(coor rhs) const //计算叉积（向量积）
    {
        return x*rhs.y - y*rhs.x;
    }
    coor operator -(coor rhs) const //坐标相减，a-b得到向量ba
    {
        return coor(x-rhs.x,y-rhs.y);
    }
    double operator *(coor rhs) const //数量积
    {
        return x*rhs.x + y*rhs.y;
    }
}a,b,liu[maxn];
struct Line
{
    coor point1,point2;
    Line(){}
    Line(coor xx,coor yy):point1(xx),point2(yy){}
}line[maxn],seg;
const double eps = 1e-14;
bool same (double a,double b)
{
    return fabs(a-b)<eps;
}
bool OnSegment (coor a,coor b,coor c) //判断点C是否在线段ab上
{
    double min_x = min(a.x,b.x), min_y = min(a.y,b.y);
    double max_x = max(a.x,b.x), max_y = max(a.y,b.y);
    if (c.x>=min_x && c.x<=max_x && c.y>=min_y && c.y<=max_y) return true;
    else return false;
}
bool SegmentIntersect (coor a,coor b,coor c,coor d)
{
    double d1 = (b-a)^(d-a); //direction(a,b,d);以a为起点，计算ab和ab的叉积
    double d2 = (b-a)^(c-a);
    double d3 = (d-c)^(a-c);
    double d4 = (d-c)^(b-c);
    if (d1*d2<0 && d3*d4<0) return true;
    else if (same(d1,0) && OnSegment(a,b,d)) return true;
    else if (same(d2,0) && OnSegment(a,b,c)) return true;
    else if (same(d3,0) && OnSegment(c,d,a)) return true;
    else if (same(d4,0) && OnSegment(c,d,b)) return true;
    else return false;
}
bool PointInPolygon (coor p[],int n,coor cmp)
{
    //思路：求解y=cmp.y与多边形一侧有多少个交点，奇数就在里面，偶数就在外面，cmp在边上是不行的
    int cnt = 0; //记录单侧有多少个交点，这里的p[]，必须有顺序
    for (int i=1;i<=n;++i)
    {
        int t = (i+1)>n ? 1:(i+1);  //下标为1要这样
        coor p1=p[i],p2=p[t];
        if (cmp.y >= max(p1.y,p2.y)) continue;//交点在延长线上和在凸顶点都不要
        if (cmp.y <  min(p1.y,p2.y)) continue;//交点在凹顶点上就要，这里没取等
        if (same(p1.y,p2.y)) continue; //与cmp.y是平行的
        double x = (cmp.y-p1.y)*(p1.x-p2.x)/(p1.y-p2.y) + p1.x; //求交点 p1.y != p2.y不会除0错误
        if (x>cmp.x) cnt++;//只统计一侧的交点
    }
    return cnt&1;
}
void work ()
{
    scanf("%lf%lf%lf%lf",&seg.point1.x,&seg.point1.y,&seg.point2.x,&seg.point2.y);
    scanf("%lf%lf%lf%lf",&a.x,&a.y,&b.x,&b.y);
    if (a.x > b.x) swap(a.x,b.x);
    if (a.y < b.y) swap(a.y,b.y);
    line[1] = Line(a,coor(b.x,a.y));
    line[2] = Line(coor(a.x,b.y),b);
    line[3] = Line(a,coor(a.x,b.y));
    line[4] = Line(coor(b.x,a.y),b);

    liu[1]=a;
    liu[2]=coor(b.x,a.y);
    liu[3]=b;
    liu[4]=coor(a.x,b.y);

    for (int i=1;i<=4;++i)
    {
        if (SegmentIntersect(seg.point1,seg.point2,line[i].point1,line[i].point2))
        {
            printf ("T\n");
            return ;
        }
    }
    if (PointInPolygon(liu,4,seg.point1)&&PointInPolygon(liu,4,seg.point2))
    {
        printf ("T\n");
        return ;
    }
    printf ("F\n");
    return ;
}

int main()
{
#ifdef local
    freopen("data.txt","r",stdin);
#endif
    int t;
    cin>>t;
    while(t--) work();
    return 0;
}