6B - President's Office

B. President's Office

time limit per test
2 seconds

memory limit per test
64 megabytes

input
standard input

output
standard output

President of Berland has a very vast office-room, where, apart from him, work his subordinates. Each subordinate, as well as President himself, has his own desk of a unique colour. Each desk is rectangular, and its sides are parallel to the office walls. One
day President decided to establish an assembly, of which all his deputies will be members. Unfortunately, he does not remember the exact amount of his deputies, but he remembers that the desk of each his deputy is adjacent to his own desk, that is to say,
the two desks (President's and each deputy's) have a common side of a positive length.

The office-room plan can be viewed as a matrix with n rows and m columns.
Each cell of this matrix is either empty, or contains a part of a desk. An uppercase Latin letter stands for each desk colour. The «period» character («.»)
stands for an empty cell.

Input

The first line contains two separated by a space integer numbers n, m (1 ≤ n, m ≤ 100)
— the length and the width of the office-room, and ccharacter — the President's desk colour. The following n lines
contain m characters each — the office-room description. It is guaranteed that the colour of each desk is unique, and each desk represents
a continuous subrectangle of the given matrix. All colours are marked by uppercase Latin letters.

Output

Print the only number — the amount of President's deputies.

Examples

input

3 4 R
G.B.
.RR.
TTT.

output

2

input

3 3 Z
...
.H.
..Z

output

0

给出一个二维数组，每个字母表示一种颜色，问指定的颜色周围有几种不同的颜色，已知相同的颜色一定有公共边

直接暴力遍历就行，注意标记遇到过的颜色

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int n,m,vis[105][105],ans=0;
char map[105][105];
int dx[]={0,-1,0,1},dy[]={1,0,-1,0};
int s[105];
void dfs(int x,int y,char ch)
{
vis[x][y]=1;
if(map[x][y]!=ch)
{
if(!s[map[x][y]])
{
++ans;
}
s[map[x][y]]=1;
return;
}
for(int i=0;i<4;++i)
{
int tx=x+dx[i],ty=y+dy[i];
if(tx<0||tx>=n||ty<0||ty>=m||map[tx][ty]=='.'||s[map[tx][ty]])
{
continue;
}
if(!vis[tx][ty])
{
dfs(tx,ty,ch);
}
}
}
void slove(char x)
{
for(int i=0;i<n;++i)
{
for(int j=0;j<m;++j)
{
if(map[i][j]==x)
{
dfs(i,j,x);
return;
}
}
}
}
int main()
{
char x;int bx,by;
scanf("%d%d %c",&n,&m,&x);
for(int i=0;i<n;++i)
{
scanf("%s",map[i]);
}
slove(x);
printf("%d\n",ans);
return 0;
}