USOPEN
OOMABO
MOOMXO
PQMROM

Being cows, their only word of interest is "MOO", which can appear in the word finder in many places, either horizontally, vertically, or diagonally. The example above contains 6 MOOs.

Farmer John is also a fan of word puzzles. Since the cows don't want him to solve their word finder before they have a chance to try it, they have encrypted its contents using a "substitution cipher" that replaces each letter of the alphabet with some different letter. For example, A might map to X, B might map to A, and so on. No letter maps to itself, and no two letters map to the same letter (since otherwise decryption would be ambiguous).

Unfortunately, the cows have lost track of the substitution cipher needed to decrypt their puzzle. Please help them determine the maximum possible number of MOOs that could exist in the puzzle for an appropriate choice of substitution cipher.

This is the same puzzle at the beginning of the problem statement after a cipher has been applied. Here "M" and "O" have been replaced with "Q" and "M" respectively.

分析：枚举每个点，对每个点，枚举他的8个方向，注意起点不能是M，终点不能是O了；

代码：

#include <bits/stdc++.h>
#define ll long long
const int maxn=1e5+10;
using namespace std;
int n,m,k,t,ma,p[300][300];
char a[51][51];
int dis[][2]={0,1,0,-1,1,0,-1,0,1,-1,1,1,-1,1,-1,-1};
void check(int x,int y)
{for(int i=0;i<8;i++){int s[2],t[2];s[0]=x+dis[i][0];s[1]=x+dis[i][0]*2;t[0]=y+dis[i][1];t[1]=y+dis[i][1]*2;if(s[1]>=0&&s[1]<n&&t[1]>=0&&t[1]<m&&a[x][y]!=a[s[0]][t[0]]&&a[s[0]][t[0]]==a[s[1]][t[1]]&&a[x][y]!='M'&&a[s[0]][t[0]]!='O')ma=max(ma,++p[a[x][y]][a[s[0]][t[0]]] );}
}
int main()
{int i,j;scanf("%d%d",&n,&m);for(i=0;i<n;i++)scanf("%s",a[i]);for(i=0;i<n;i++)for(j=0;j<m;j++){check(i,j);}printf("%d\n",ma);//system("pause");return 0;
}