You are given a point set with nn points on the 2D-plane, your task is to find the smallest number of points you need to add to the point set, so that all the points in the set are center symmetric.
All the points are center symmetric means that you can find a center point (X_c,Y_c)(Xc,Yc)(not necessarily in the point set), so that for every point (X_i,Y_i)(Xi,Yi) in the set, there exists a point (X_j,Y_j)(Xj,Yj) (ii can be equal to jj) in the set satisfying X_c=(X_i X_j)/2Xc=(Xi Xj)/2 and Y_c=(Y_i Y_j)/2Yc=(Yi Yj)/2.
Input
The first line contains an integer n(1 le n le 1000)n(1≤n≤1000).
The next nn lines contain nn pair of integers (X_i,Y_i)(Xi,Yi) (-10^6 le X_i,Y_i le 10^6)(−106≤Xi,Yi≤106) -- the points in the set
Output
Output a single integer -- the minimal number of points you need to add.
样例输入复制
代码语言：javascript复制3
2 0
-3 1
0 -2样例输出复制
代码语言：javascript复制1样例解释
For sample 11, add point (5,-3)(5,−3) into the set, the center point can be (1,-1)(1,−1) .
这个题，卡精度，如果除2之后，C 编译器抽风的出来的值不一样，存map就不准了，所以不除2，直接相加储存。
挺简单的。
代码语言：javascript复制#include<cstdio>
#include<vector>
#include<queue>
#include<cmath>
#include<iostream>
#include<cstring>
#include<map>
#include<algorithm>
using namespace std;
#define LL long long
#define MAXN 1100
#define esp 1e-2
map<pair<long long ,long long >,int>m;
int x[MAXN],y[MAXN];
int main()
{
double a,b;
int n,i,j,maxx=0;
scanf("%d",&n);
if(n==1||n==2) return cout<<0<<endl,0;
for(i=1; i<=n; i )
scanf("%d%d",&x[i],&y[i]);
for(i=1; i<=n; i )
for(j=i 1; j<=n; j )
{
a=x[i] x[j];
b=y[i] y[j];
m[make_pair(a,b)] ;
maxx=max(maxx,m[make_pair(a,b)]);
}
int cnt=0;
for(auto po=m.begin(); po!=m.end(); po )
{
if(po->second==maxx)
{
int a=po->first.first;
int b=po->first.second;
int w=0;
for(i=1; i<=n; i )
{
if(a==2*x[i]&&b==2*y[i]) w ;
}
cnt=max(cnt,w);
}
}
printf("%d",n-2*maxx-cnt);
return 0;
}签到题，以为很难，没做。
