传送门：HDU4739
【题意】给出N个点，求出这些点能组成的平行于坐标轴的正方形的点数。
【分析】坑点是有重点，所以我这里直接按值记录所有点出现次数和正方形，首先枚举正方形的四个点，把所有能组成的不同正方形记录下来，然后再从这些正方形中使用四个点最小剩余次数的点数作为该正方形的次数，组成后4个点都要减去相应次数，然后就好了。复杂度n^4,对于再大点的n，比状压的2^n要快很多。
【AC CODE】15ms
#include <cstdio>
#include <cstring>
#include <cctype>
#include <cmath>
#include <map>
#include <set>
//#include <unordered_set>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <algorithm>
using namespace std;
typedef long long LL;
#define rep(i,a,n) for(int i = a; i < n; i++)
#define repe(i,a,n) for(int i = a; i <= n; i++)
#define per(i,n,a) for(int i = n; i >= a; i--)
#define clc(a,b) memset(a,b,sizeof(a))
const int INF = 0x3f3f3f3f, MAXN = 20+10;
struct NODE{
int x, y;
bool operator<(const NODE &t) const{
if(x != t.x) return x < t.x;
return y < t.y;
}
bool operator!=(const NODE &t)const{
return !(x==t.x && y == t.y);
}
};
struct SQU{
NODE p[4];
bool operator<(const SQU&t) const{
rep(i,0,4) if(p[i] != t.p[i]) return p[i] < t.p[i];
return false;
}
};
int x[MAXN],y[MAXN];
int a[4];
bool cmp(int p, int q)
{
if(x[p] != x[q]) return x[p] < x[q];
return y[p] < y[q];
}
bool ok(int aa, int b, int c, int d)
{
a[0] = aa,a[1] = b,a[2] = c, a[3] = d;
sort(a,a+4,cmp);
rep(i,1,4)
{
if(x[a[i]] == x[a[i-1]] && y[a[i]] == y[a[i-1]])
return false;
}
set<int> fx,fy;
int cntx = 0,cnty = 0;
rep(i,0,4)
{
fx.insert(x[a[i]]);
fy.insert(y[a[i]]);
}
if(fx.size() != 2 || fy.size() != 2) return false;
auto it = fx.begin();
int tmp = *(it);it++;
int ans1 = (*it)-tmp;
it = fy.begin();
tmp = *(it);it++;
int ans2 = (*it)-tmp;
return ans1 == ans2;
}
int tol[MAXN];
int main()
{
#ifdef SHY
freopen("d:\\1.txt", "r", stdin);
#endif
int n;
while(~scanf("%d", &n) && ~n)
{
map<NODE,int> cnt;
clc(tol,0);
rep(i,0,n)
{
scanf("%d %d", &x[i], &y[i]);
cnt[NODE{x[i],y[i]}]++;
}
set<SQU> all;
rep(i,0,n)
{
rep(j,i+1,n)
{
rep(k,j+1,n)
{
rep(l,k+1,n)
{
if(ok(i,j,k,l))
{
NODE a = {x[i],y[i]},b = {x[j],y[j]},c = {x[k],y[k]},d = {x[l],y[l]};
SQU tmp = {a,b,c,d};
sort(tmp.p,tmp.p+4);
all.insert(tmp);
}
}
}
}
}
int ans = 0;
for(auto it = all.begin(); it != all.end(); it++)
{
int mi = INF;
rep(i,0,4)
{
int tmp = cnt[it->p[i]];
if(mi > tmp) mi = tmp;
}
rep(i,0,4)
cnt[it->p[i]] -= mi;
ans += mi;
}
printf("%d\n", ans*4);
}
return 0;
}
/*
18
3 2
3 0
4 3
0 0
1 1
1 0
4 0
0 3
0 2
3 3
1 2
1 2
2 4
2 1
1 0
3 1
1 3
1 2
19
0 2
2 1
0 1
4 3
1 3
4 4
0 2
1 0
3 1
0 3
4 0
3 1
0 3
1 4
1 4
1 3
1 2
1 0
4 3
18
3 2
3 0
4 3
0 0
1 1
1 0
4 0
0 3
0 2
3 3
1 2
1 2
2 4
2 1
1 0
3 1
1 3
1 2
8
0 0
0 1
1 0
1 1
1 0
1 1
2 0
2 1
20
0 0
0 1
1 0
1 1
0 0
0 1
1 0
1 1
0 0
0 1
1 0
1 1
0 0
0 1
1 0
1 1
0 0
0 1
1 0
1 1
3
1 1
0 0
2 2
8
0 0
1 0
2 0
0 1
1 1
2 1
10 1
10 0
-1
*/