[1443] Weiqi

• [1443] Weiqi

• http://ac.nbutoj.com/Problem/view.xhtml?id=1443
• 时间限制: 1000 ms　内存限制: 65535 K
• 问题描述
• Have you ever played Weiqi?
  In a composition, there exist two kinds of chess pieces black and white.
  The problem is so easy, can you find out how many white pieces are encircled by blacks? 
  
  
• 输入
• First line will contain two integers M and N (3 <= M, N <= 100), means the size of the composition.
  Then M lines, each line N integers only including '1', '2', and '0' ('1' represent black piece, '2' represent white piece and '0' represent empty).
• 输出
• Print how many white pieces are encircled by black. 
• 样例输入

• 4 4
  
  0012
  
  0121
  
  0111
  
  0101
  
  4 4
  
  0012
  
  0121
  
  0101
  
  0111
  
  

• 样例输出

• 1
  
  0
  
  

• 提示

• 详情看样例输入输出

• 来源

• Hungar

• 操作

 

 

#include<iostream>

#include<cstdio>

#include<cstring>



using namespace std;



char map[110][110];

int vis[110][110];



int m,n,flag,tag;



int dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}};



int cnt;



void DFS(int x,int y){

    for(int i=0;i<4;i++){

        int sx=x+dir[i][0];

        int sy=y+dir[i][1];

        if(sx<0 || sx>=m || sy<0 || sy>=n || map[sx][sy]=='1')

            continue;

        if(map[sx][sy]=='0' || sx==0 || sx==m-1 || sy==0 || sy==n-1)

            tag=1;

        //printf("22222\n");

        //printf("x=%d y=%d\n",x,y);

        //printf("sx=%d sy=%d\n",sx,sy);

        //printf("------- cnt=%d\n",cnt);

        if(!vis[sx][sy]){

            vis[sx][sy]=1;

            cnt++;

            DFS(sx,sy);

        }

    }

}



int main(){



    //freopen("input.txt","r",stdin);



    while(~scanf("%d%d",&m,&n)){

        getchar();

        for(int i=0;i<m;i++){

            for(int j=0;j<n;j++)

                scanf("%c",&map[i][j]);

            getchar();

        }

        memset(vis,0,sizeof(vis));

        int ans=0;

        for(int i=1;i<m-1;i++)

            for(int j=1;j<n-1;j++){

                if(map[i][j]=='2' && !vis[i][j]){

                    //printf("i=%d j=%d\n",i,j);

                    vis[i][j]=1;

                    cnt=1;

                    tag=0;

                    DFS(i,j);

                    //printf("cnt=%d\n",cnt);

                    if(!tag)

                        ans+=cnt;

                }

            }

        printf("%d\n",ans);

    }

    return 0;

}