AcWing 1749. Block billboard II (classified discussion + enumeration)

subject ：1749. Block billboards II

Answer key ： The last part not covered , Project to x Axis and y The length on the shaft , Multiplication is the answer . Equivalent to finding... In the uncovered part x Maximum on shaft 、 Small value .y So is the shaft . Multiply the two . Note that if it's covered , You can't find the maximum or minimum value , So remember and 0 Compare

#include<bits/stdc++.h>
#define x first
#define y second
using namespace std;
typedef long long LL;
typedef pair<int ,int> PII;
const int N=1e5+10;
const int mod=100000007;
bool sta[2050][2050];

int main(){
    
    int a[3][4];
    int sum=0;
    for(int i=0;i<2;i++){
    
        cin>>a[i][0]>>a[i][1]>>a[i][2]>>a[i][3];
    }
    for(int k=0;k<2;k++){
    
        for(int i=a[k][0]+1000;i<a[k][2]+1000;i++){
    
            for(int j=a[k][1]+1000;j<a[k][3]+1000;j++){
    
                if(k==0){
    
                    sta[i][j]=1;
                }else{
    
                    sta[i][j]=0;
                }
            }
        }
    }
    int b=2050,c=2050,d=0,e=0;//b、d Record x Minimum on shaft 、 Great value .c,e Record y Minimum on shaft 、 Great value 
    for(int i=a[0][0]+1000;i<a[0][2]+1000;i++){
    
        for(int j=a[0][1]+1000;j<a[0][3]+1000;j++){
    
            if(sta[i][j]){
    
                b=min(b,i),d=max(d,i);
                c=min(c,j),e=max(e,j);
            }
        }
    }
    cout<<max(0,d-b+1)*max(0,e-c+1);// Pay attention to the situation where they are covered 
    return 0;
}