B - Calabash and Landlord 解题报告

题目链接：https://vjudge.net/contest/381753#problem/B
解题报告：
题目大意：
两个矩形能把平面分为多少个区域？
解题思路：
列举两个矩形的关系。别漏，列全~

代码：

#include<bits/stdc++.h>
using namespace std;
bool check(int x1, int y1, int x2, int y2, int x3, int y3, int x4, int y4)
{
    return (x1<=x3&&y1<=y3&&x4<=x2&&y4<=y2);
}
bool check1(int x1, int y1, int x2, int y2, int x3, int y3, int x4, int y4)
{
    return (x3>x1&&y3<y1&&x4<x2&&y4>y2);
}
bool check2(int x1, int y1, int x2, int y2, int x3, int y3, int x4, int y4)
{
    return ( (x3==x1&&y3<y1&&x4<x2&&y4>y2) || (x4==x2&&y3<y1&&x3>x1&&y4>y2) || (x3>x1&&y3==y1&&x4<x2&&y4>y2) || (x3>x1&&y3<y1&&x4<x2&&y4==y2) );
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--) {
        int ans;
        int x1, y1, x2, y2, x3, y3, x4, y4;
        scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
        scanf("%d%d%d%d",&x3,&y3,&x4,&y4);
        if(x1==x3&&x2==x4&&y1==y3&&y2==y4) {    //重合 
            ans = 2;
        }
        else if(x2<=x3||x1>=x4||y2<=y3||y1>=y4) {    //相离 
            ans = 3;
        }
        else if(check(x1,y1,x2,y2,x3,y3,x4,y4) || check(x3,y3,x4,y4,x1,y1,x2,y2)) {    //包含 
            if((x1==x3&&x2==x4&&y1!=y3&&y2!=y4) || (y1==y3&&y2==y4&&x1!=x3&&x2!=x4)) {    //其中一类情况 
                ans = 4;
            }            
            else {
                ans = 3;
            }
        }
        else {
            if(check1(x1,y1,x2,y2,x3,y3,x4,y4) || check1(x3,y3,x4,y4,x1,y1,x2,y2)) {    //相交，别忘了是对称的 
                ans = 6;
            }
            else if(check2(x1,y1,x2,y2,x3,y3,x4,y4) || check2(x3,y3,x4,y4,x1,y1,x2,y2)) {    //其中一类情况 
                ans = 5;
            }
            else {
                ans = 4;
            }
        }
        printf("%d\n",ans);
    }
}