判断矩形相交的方法

给你两个矩形，分别给你矩形的左下角和右上角，问你是否相交。

第一个矩形：（x1,y1),(x2,y2)
第二个矩形：  (x3,y3),(x4,y4)

如果满足max（x1,x3）<=min(x2,x4)&&max(y1,y3)<=min(y2,y4)，则相交。

具体链接：http://codeforces.com/contest/1080/problem/C

AC代码：

#include <bits/stdc++.h>
using namespace std;
# define ll long long
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        ll n,m;
        scanf("%lld %lld",&n,&m);
        ll x1,y1,x2,y2;
        ll x3,y3,x4,y4;
        scanf("%lld %lld %lld %lld %lld %lld %lld %lld",&x1,&y1,&x2,&y2,&x3,&y3,&x4,&y4);
        ll white=0;
        // if(n%2!=0&&m%2!=0)
        // {
        white=(n*m+1)/2;
       // cout<<1<<" "<<white<<endl;
        // }
        // else
        //    white=(n*m)/2;
        // int len1,len2;
        if((x1%2+y1%2)==0||(x1%2+y1%2)==2)
        {
            white+=((x2-x1+1)*(y2-y1+1))/2;
        }
        else
            white+=((x2-x1+1)*(y2-y1+1)+1)/2;
         //cout<<2<<" "<<white<<endl;
        if((x3%2+y3%2)==0||(x3%2+y3%2)==2)
        {
            white-=((x4-x3+1)*(y4-y3+1)+1)/2;
        }
        else
            white-=((x4-x3+1)*(y4-y3+1))/2;
         //cout<<3<<" "<<white<<endl;
        ll x5,y5,x6,y6;
        x5=max(x1,x3);
        x6=min(x2,x4);
        y5=max(y1,y3);
        y6=min(y2,y4);
       // cout<<white<<endl;
        //cout<<x5<<" "<<y5<<endl<<x6<<" "<<y6<<endl;
        if(x5<=x6&&y5<=y6)
        {
            if((x5%2+y5%2)==0||(x5%2+y5%2)==2)
            {
                white-=((x6-x5+1)*(y6-y5+1))/2;
            }
            else
                white-=((x6-x5+1)*(y6-y5+1)+1)/2;
        }
       // cout<<white<<endl;
        printf("%lld %lld\n",white,n*m-white);
    }
    return 0;
}