2019ACM-ICPC徐州网络赛-K- Center（简单贪心+map）-优快云博客

本文探讨了在二维平面上，如何通过增加最少数量的点，使得给定点集中的所有点都成为中心对称。中心对称意味着存在一个中心点，使得每一点都有一个对应的点满足中心对称条件。文章提供了算法思路和实现代码。

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K- Center

2000ms 262144K

You are given a point set with

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 $\le n \le 1000)n(1≤n≤1000).$

The next $(X_i,Y_i)(Xi,Yi) (-10^6 \le X_i,Y_i \le 10^6)(−106≤Xi,Yi≤106) -- the points in the set$

Output

Output a single integer -- the minimal number of points you need to add.

样例输入

样例输出

样例解释

For sample

 1 #include <bits/stdc++.h>
 2 #pragma GCC optimize(3)
 3 using namespace std;
 4 typedef long long ll;
 5 typedef pair<int,int> PII;
 6 const int maxn=1e6+7;
 7 map<PII,int>MAP;
 8 int x[maxn],y[maxn];
 9 int main()
10 {
11     int n;
12     int ans=0;
13     scanf("%d",&n);
14     for(int i=1;i<=n;++i)
15     {
16         scanf("%d%d",&x[i],&y[i]);
17         x[i]*=2;
18         y[i]*=2;
19     }
20     for(int i=1;i<=n;++i)
21     {
22         for(int j=1;j<=n;++j)
23         {
24             int a=++MAP[PII((x[i]+x[j])/2,(y[i]+y[j])/2)];
25             if(a>ans)ans=a;
26         }
27     }
28     printf("%d\n",n-ans);
29     return 0;
30 }