A
little known fact about cows is the fact that they are red-green
colorblind, meaning that red and green look identical to them.  This
makes it especially difficult to design artwork that is appealing to
cows as well as humans.

Consider a square painting that is
described by an N x N grid of characters (1 <= N <= 100), each one
either R (red), G (green), or B (blue).  A painting is interesting if
it has many colored "regions" that can be distinguished from
each-other.  Two characters belong to the same region if they are
directly adjacent (east, west, north, or south), and if they are
indistinguishable in color.  For example, the painting
RRRBB
GGBBB
BBBRR
BBRRR
RRRRR
has 4 regions (2 red, 1 blue, and 1 green) if viewed by a human, but only 3 regions (2 red-green, 1 blue) if viewed by a cow.

Given a painting as input, please help compute the number of regions in the painting when viewed by a human and by a cow.

* Line 1: The integer N.
* Lines 2..1+N: Each line contains a string with N characters,describing one row of a painting.

* Line 1: Two space-separated integers, telling the number of regions in the painting when viewed by a human and by a cow.

5

RRRBB

GGBBB

BBBRR

BBRRR

RRRRR

4 3
【分析】水题，两次BFS。

#include <iostream>

#include <cstdio>

#include <cstring>

#include <set>

#include <queue>

#include <algorithm>

#define MAXN 111111

#define MAXM 222222

#define INF 1000000000

using namespace std;

const int N=1e2+;

int cnt,rt,n;

int vis[N][N];

int d[][]={,,,,-,,,-};

char str[N][N];

struct man{

    int x,y;

};

void bfs(int x,int y){

    man s;s.x=x;s.y=y;

    queue<man>q;

    q.push(s);

    vis[x][y]=;

    while(!q.empty()){

        man t=q.front();

        q.pop();

        for(int i=;i<;i++){

            int xx=t.x+d[i][];

            int yy=t.y+d[i][];

            if(xx>=&&xx<n&&yy>=&&yy<=n&&!vis[xx][yy]&&str[xx][yy]==str[t.x][t.y]){

                man k;k.x=xx;k.y=yy;

                q.push(k);

                vis[xx][yy]=;

            }

        }

    }

}

int main(){

    scanf("%d",&n);

    int ans1=,ans2=;

    for(int i=;i<n;i++)scanf("%s",str[i]);

    for(int i=;i<n;i++){

        for(int j=;j<n;j++)

           if(!vis[i][j]){

              bfs(i,j);

              ans1++;

        }

    }

    for(int i=;i<n;i++){

        for(int j=;j<n;j++){

           if(str[i][j]=='R')str[i][j]='G';

        }

    }

    memset(vis,,sizeof(vis));

    for(int i=;i<n;i++){

        for(int j=;j<n;j++)

           if(!vis[i][j]){

              bfs(i,j);

              ans2++;

        }

    }

    printf("%d %d\n",ans1,ans2);

    return ;

}