Intersection

题意：判断一条线段是否在矩形里或与矩形相交。

思路：先判断线段的两个点是否在矩形里，再判断线段是否与矩形的四条边相交即可。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<iostream>
using namespace std;
typedef long long ll;
ll chaj(ll x1,ll y1,ll x2,ll y2)
{
    ll ans=x1*y2-x2*y1;
    if(ans==0)
        return 0;
    if(ans>0)
        return 1;
    return -1;
}
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        ll x1,y1,x2,y2,x3,y3,x4,y4;
        cin>>x1>>y1>>x2>>y2>>x3>>y3>>x4>>y4;
        if(x1>=min(x3,x4)&&x1<=max(x3,x4)&&y1>=min(y3,y4)&&y1<=min(y3,y4))
        {
            cout<<"T"<<endl;
            continue;
        }
        if(x2>=min(x3,x4)&&x2<=max(x3,x4)&&y2>=min(y3,y4)&&y2<=min(y3,y4))
        {
            cout<<"T"<<endl;
            continue;
        }
        int s=0;
        if( chaj(x1-x3,y1-y3,0,y4-y3)*chaj(x2-x3,y2-y3,0,y4-y3)<=0&&chaj(x3-x1,y3-y1,x2-x1,y2-y1)*chaj(x3-x1,y4-y1,x2-x1,y2-y1)<=0  )
            s=1;
        if( chaj(x1-x3,y1-y3,x4-x3,0)*chaj(x2-x3,y2-y3,x4-x3,0)<=0&&chaj(x3-x1,y3-y1,x2-x1,y2-y1)*chaj(x4-x1,y3-y1,x2-x1,y2-y1)<=0  )
            s=1;
        if( chaj(x1-x4,y1-y4,x3-x4,0)*chaj(x2-x4,y2-y4,x3-x4,0)<=0&&chaj(x4-x1,y4-y1,x2-x1,y2-y1)*chaj(x3-x1,y4-y1,x2-x1,y2-y1)<=0  )
            s=1;
        if( chaj(x1-x4,y1-y4,0,y3-y4)*chaj(x2-x4,y2-y4,0,y3-y4)<=0&&chaj(x4-x1,y4-y1,x2-x1,y2-y1)*chaj(x4-x1,y3-y1,x2-x1,y2-y1)<=0  )
            s=1;
        if(s)
            cout<<"T"<<endl;
        else
            cout<<"F"<<endl;
    }
}