POJ 1410 Intersection（计算几何）

题目大意：题目意思很简单，就是说有一个矩阵是实心的，给出一条线段，问线段和矩阵是否相交
解题思路：用到了线段与线段是否交叉，然后再判断线段是否在矩阵里面，这里要注意的是，他给出的矩阵的坐标明显不是左上和右下的坐标，需要自己去判断下左上点与右下点的坐标。

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

const int INF=0x3f3f3f3f;
typedef long long ll;
#define prN printf("
")
#define SI(N) scanf("%d",&(N))
#define SII(N,M) scanf("%d%d",&(N),&(M))
#define SIII(N,M,K) scanf("%d%d%d",&(N),&(M),&(K))
#define cle(a,val) memset(a,(val),sizeof(a))
#define rep(i,b) for(int i=0;i<(b);i++)
#define Rep(i,a,b) for(int i=(a);i<=(b);i++)
#define reRep(i,a,b) for(int i=(a);i>=(b);i--)
const double eps = 1e-8;
//判断doubule型的正负或0
int sgn(double x)
{
    if(fabs(x) < eps)return 0;
    if(x < 0) return -1;
    else return 1;
}
//构建点，且重载运算符
struct Point
{
    double x,y;
    Point(){}
    Point(double _x,double _y)
    {
        x = _x;y = _y;
    }
    //重载减号 因为在求两个点相减构成一个向量时候会用到
    Point operator -(const Point &b)const
    {
        return Point(x - b.x,y - b.y);
    }
    //这是叉积运算，很重要，不多说
    double operator ^(const Point &b)const
    {
        return x*b.y - y*b.x;
    }
    double operator *(const Point &b)const
    {
        return x*b.x + y*b.y;
    }
};
struct Line
{
    Point s,e;
    Line(){}
    Line(Point _s,Point _e)
    {
        s = _s;e = _e;
    }
};
//判断线段相交
bool inter(Line l1,Line l2)
{
    return
        //这是2个矩形是否相交
        max(l1.s.x,l1.e.x) >= min(l2.s.x,l2.e.x) &&
        max(l2.s.x,l2.e.x) >= min(l1.s.x,l1.e.x) &&
        max(l1.s.y,l1.e.y) >= min(l2.s.y,l2.e.y) &&
        max(l2.s.y,l2.e.y) >= min(l1.s.y,l1.e.y) &&
        //这是判断叉积异号
        sgn((l2.s-l1.s)^(l1.e-l1.s))*sgn((l2.e-l1.s)^(l1.e-l1.s)) <= 0 &&
        sgn((l1.s-l2.s)^(l2.e-l2.s))*sgn((l1.e-l2.s)^(l2.e-l2.s)) <= 0;
}
//求距离
double dist(Point a,Point b)
{
    return sqrt((b-a)*(b-a));
}

int n;
Line line[4];
Line endli;
double a1,a2,a3,a4,xtop,ytop,xbott,ybott;
int main()
{
#ifndef ONLINE_JUDGE
    freopen("C:\Users\Zmy\Desktop\in.txt","r",stdin);
//    freopen("C:\Users\Zmy\Desktop\out.txt","w",stdout);
#endif // ONLINE_JUDGE
    SI(n);
    rep(t,n)
    {
        scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&a1,&a2,&a3,&a4,&xtop,&ytop,&xbott,&ybott);
        endli=Line(Point(a1,a2),Point(a3,a4));
        //把4条边存进line里
        line[0]=Line(Point(xtop,ytop),Point(xtop,ybott));
        line[1]=Line(Point(xtop,ytop),Point(xbott,ytop));
        line[2]=Line(Point(xtop,ybott),Point(xbott,ybott));
        line[3]=Line(Point(xbott,ytop),Point(xbott,ybott));
        int fl=0;
        //判断两条线是否相交
        rep(i,4)
        {
            if (inter(line[i],endli))
            {
                fl=1;break;
            }
        }
        //判断是否在矩形内
        if (max(a1,a3)<max(xbott,xtop)&&min(a1,a3)>min(xbott,xtop)&&max(a2,a4)<max(ytop,ybott)&&min(a2,a4)>min(ytop,ybott))
        {
            fl=1;
        }
        if (fl) puts("T");
        else puts("F");
    }
    return 0;
}