[1443] Weiqi-白红宇

[1443] Weiqi

阅读量：6758 次

发布时间：2019-06-26

本文共 1514 字，大约阅读时间需要 5 分钟。

[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 400120121011101014 40012012101010111
```

样例输出

```
10
```

提示

```
详情看样例输入输出
```

来源

```
Hungar
```

操作

#include
     
      #include
      
       #include
       
        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