The Preliminary Contest for ICPC Asia Xuzhou 2019 K. Center

Topic links: https://nanti.jisuanke.com/t/41393

Title effect: analysis n points, plus a minimum number of points Q such that the center point of symmetry.

answer:

Two point may determine a center point, O (n ^ 2) enumeration central point, provided cnt [Xc] [Yc] of the center point (Xc, Yc) are pieces

For the number of times,

Then the center point (Xc, Yc), the number of points required to make up the n-2 * cnt [Xc] [Yc] - (point (Xc, Yc) in the original

The number of points in the set appears),

You can map or a hash value cnt recording.

In addition, if the input coordinates are multiplied by 2, then to enumerate the coordinates of the center point on all the integers.

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
struct st{
    ll x,y;
}stm[1005];
map<ll,int> mp;
map<ll,int> mp1;
set<ll> se;
const ll tem=1e7;
queue<ll> que;
map<ll,int> vis;
int main(){
    int n;
    scanf("%d",&n);
    int maxs=0;
    for(int i=0;i<n;i++){
        scanf("%lld%lld",&stm[i].x,&stm[i].y);
        stm[i].x*=2;
        stm[i].y*=2;
        ll tt=stm[i].x*tem+stm[i].y;
        mp[tt]++;
    }
    for(int i=0;i<n;i++){
        for(int j=0;j<n;j++){
            if(i<j){
                ll tt=(stm[i].x+stm[j].x)/2*tem+(stm[i].y+stm[j].y)/2;
                int count=++mp1[tt];
                if(count>maxs){
                    maxs=count;
                    while(!que.empty())que.pop();
                    que.push(tt);
                }
                else if(count==maxs){
                    que.push(tt);
                }
            }      
        }
    }
    int= ANS n-; 
   
    the while (! que.empty ()) { 
        LL TT = que.front (); 
        que.pop (); 
        ANS = min (ANS, n--MP1 [TT] * 2 - (MP [TT] & . 1 ));
         IF (MP [TT] & . 1 ) BREAK ;
     //     COUT TT << << endl; 
    }
     for ( int I = 0 ; I <n-; I ++) { // consider the case where the center point at the origin 
        ll STM = TT [I] .x + * TEM STM [I] .y; 
        ANS = min (ANS, n--MP1 [TT] * 2 - (MP [TT] & . 1 )); 
    } 
    the printf ( "%d\n",ans);
    return 0;
}