HDU - 5128The E-pang Palace+暴力枚举，计算几何

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

#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#define pb push_back
#define pq priority_queue

#define Pll pair<ll,ll>
#define Pii pair<int,int>

#define fi first
#define se second

#define OKC ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
typedef long long ll;
typedef unsigned long long ull;


/*-----------------show time----------------*/
int n;
int l[40],r[40];
int mp[205][205];
bool check(int i,int j,int x,int y)
{
    int e[5][4];
    e[1][1] = l[i],e[1][2] = r[i];
    e[2][1] = l[i],e[2][2] = r[j];
    e[3][1] = l[j],e[3][2] = r[j];
    e[4][1] = l[j],e[4][2] = r[i];
    int le1 = min(l[x],l[y]),le2 = max(l[x],l[y]);
    int ri1 = min(r[x],r[y]),ri2 = max(r[x],r[y]);
    if(mp[l[x]][r[y]]==0||mp[l[y]][r[x]]==0)return false;
    for(int w=1; w<=4; w++)
    {
        if(mp[e[w][1]][e[w][2]]==0)return false;
        if(e[w][1]>=le1 && e[w][1] <= le2 && e[w][2]>=ri1 && e[w][2] <= ri2)
            return false;
    }
    return true;
}
/*
bool tepan(int i,int j,int q,int w)
{
    //1
    int l1 = min(l[i],l[j]),l2 = max(l[i],l[j]);
    int r1 = min(r[i],r[j]),r2 = max(r[i],r[j]);

    //2

    int x1 = min(l[q],l[w]),x2 = max(l[q],l[w]);
    int y1 = min(r[q],r[w]),y2 = max(r[q],r[w]);

    if(y1<r1 && y2>r2 && l1<x1 && l2 > x2)return true;
    if(y1>r1 && y2<r2 && l1>x1 && l2 < x2)return true;
    return false;
}
*/
int main(){
    while(~scanf("%d", &n)&&n){
        memset(mp,0,sizeof(mp));
        for(int i=1; i<=n; i++){
            scanf("%d%d", &l[i],&r[i]);
            mp[l[i]][r[i]] = 1;
        }
        int ans = 0;
        for(int i=1; i<=n; i++){
        for(int j=1; j<=n; j++){
        for(int q = 1; q<=n; q++){
        for(int w = 1; w<=n; w++){

            if(check(i,j,q,w)&&check(q,w,i,j)){
                //if(tepan(i,j,q,w) ||tepan(q,w,i,j))continue;
                if(l[i]==l[j]||r[i]==r[j]||l[q]==l[w]||r[q]==r[w])continue;
                int tmp = abs(l[i]-l[j]) * abs(r[i] - r[j]) + abs(l[q]-l[w]) * abs(r[q] - r[w])  ;
                ans = max (ans,tmp);    
            }    
            else if(check(i,j,q,w))
            {
                if(l[i]==l[j]||r[i]==r[j]||l[q]==l[w]||r[q]==r[w])continue;
                if(mp[l[q]][r[w]]==0||mp[l[w]][r[q]]==0)continue;
                int tmp = abs(l[i]-l[j]) * abs(r[i] - r[j]) ;
                            // int tmp = max(tmp,)
                int e[5][4];
                e[1][1] = l[q],e[1][2] = r[q];
                e[2][1] = l[q],e[2][2] = r[w];
                e[3][1] = l[w],e[3][2] = r[w];
                e[4][1] = l[w],e[4][2] = r[q];
                int le1 = min(l[i],l[j]),le2 = max(l[i],l[j]);
                int ri1 = min(r[i],r[j]),ri2 = max(r[i],r[j]);
                int flag = 1;
                for(int s = 1; s<=4;s++){                        //用于判断回字型的合法，里面完全包于
                    if(e[s][1]==le1||e[s][1]==le2||e[s][2]==ri1||e[s][2]==ri1){
                        flag = 0;
                        break;
                    }
                }
                if(flag)ans = max (ans,tmp);  
                /*                            
                if(tmp==3)                                      //样例一自己跑出3.发现了回型矩阵的奥秘；
                {
                    cout<<"----------------"<<endl;
                    cout<<"zhe;"<<i<<" "<<j<<" "<<q<<" "<<w<<endl;                                                    
                    int e[5][4];
                    e[1][1] = l[i],e[1][2] = r[i];
                    e[2][1] = l[i],e[2][2] = r[j];
                    e[3][1] = l[j],e[3][2] = r[j];
                    e[4][1] = l[j],e[4][2] = r[i];
                    int le1 = min(l[q],l[w]),le2 = max(l[q],l[w]);
                    int ri1 = min(r[q],r[w]),ri2 = max(r[q],r[w]);
                    debug(le1);debug(le2);debug(ri1);debug(ri2);
                }
                */
            }

        }}}}        //这是四重循环的右括号
        if(ans == 0)puts("imp");
        else printf("%d\n",ans);
    }
    return 0;
}