记录编号	559960	评测结果	AAAAAAAAAA
题目名称	知己知彼，百战不殆	最终得分	100
用户昵称	遥时_彼方	是否通过	通过
代码语言	C++	运行时间	0.090 s
提交时间	2021-03-31 18:48:51	内存使用	10.95 MiB

#include<bits/stdc++.h>
#define ull unsigned long long
#define ll long long
using namespace std;
int n[1801][1801]; 
int nc,mc; 
int ans;
int mx;
int mj;
void ad(int x,int y)
{
    n[x][y]++;
    if(mx<n[x][y])
    {
        mx=n[x][y];
        mj=1;
    }
    else if(mx==n[x][y]) mj++;
    if(n[x][y]==1) ans++;
    if(x-2>0&&y+1<=nc)
    {
        if(!n[x-2][y+1]) ans++;
        n[x-2][y+1]++;
        if(mx<n[x-2][y+1])
        {
            mx=n[x-2][y+1];
            mj=1;
        }
        else if(mx==n[x-2][y+1]) mj++;
    }
    if(x-1>0&&y+2<=nc)
    {
        if(!n[x-1][y+2]) ans++;
        n[x-1][y+2]++;
        if(mx<n[x-1][y+2])
        {
            mx=n[x-1][y+2];
            mj=1;
        }
        else if(mx==n[x-1][y+2]) mj++;
    }
    if(x+1<=nc&&y+2<=nc)
    {
        if(!n[x+1][y+2]) ans++;
        n[x+1][y+2]++;
        if(mx<n[x+1][y+2])
        {
            mx=n[x+1][y+2];
            mj=1;
        }
        else if(mx==n[x+1][y+2]) mj++;
    }
    if(x+2<=nc&&y+1<=nc)
    {
        if(!n[x+2][y+1]) ans++;
        n[x+2][y+1]++;
        if(mx<n[x+2][y+1])
        {
            mx=n[x+2][y+1];
            mj=1;
        }
        else if(mx==n[x+2][y+1]) mj++;
    }
    if(x+1<=nc&&y-2>0)
    {
        if(!n[x+1][y-2]) ans++;
        n[x+1][y-2]++;
        if(mx<n[x+1][y-2])
        {
            mx=n[x+1][y-2];
            mj=1;
        }
        else if(mx==n[x+1][y-2]) mj++;
    }
    if(x-1>0&&y-2>0)
    {
        if(!n[x-1][y-2]) ans++;
        n[x-1][y-2]++;
        if(mx<n[x-1][y-2])
        {
            mx=n[x-1][y-2];
            mj=1;
        }
        else if(mx==n[x-1][y-2]) mj++;
    }
    if(x-2>0&&y-1>0)
    {
        if(!n[x-2][y-1]) ans++;
        n[x-2][y-1]++;
        if(mx<n[x-2][y-1])
        {
            mx=n[x-2][y-1];
            mj=1;
        }
        else if(mx==n[x-2][y-1]) mj++;
    }
    if(x+2<=nc&&y-1>0)
    {
        if(!n[x+2][y-1]) ans++;
        n[x+2][y-1]++;
        if(mx<n[x+2][y-1])
        {
            mx=n[x+2][y-1];
            mj=1;
        }
        else if(mx==n[x+2][y-1]) mj++;
    }
    return;
}
int main()
{
    freopen("safenum.in","r",stdin);
	freopen("safenum.out","w",stdout);
	cin>>nc>>mc;
	int s1,s2; 
	for(int i=1;i<=mc;i++)
	{
	    scanf("%d%d",&s1,&s2);
	    ad(s1,s2);
    }
    cout<<nc*nc-ans<<endl<<mx<<" "<<mj<<endl;
    return 0;
}