度度熊的01世界

 

 Accepts: 967

 

 Submissions: 3064

 Time Limit: 2000/1000 MS (Java/Others)

 

 Memory Limit: 32768/32768 K (Java/Others)

Problem Description

度度熊是一个喜欢计算机的孩子，在计算机的世界中，所有事物实际上都只由0和1组成。

现在给你一个n*m的图像，你需要分辨他究竟是0，还是1，或者两者均不是。

图像0的定义：存在1字符且1字符只能是由一个连通块组成，存在且仅存在一个由0字符组成的连通块完全被1所包围。

图像1的定义：存在1字符且1字符只能是由一个连通块组成，不存在任何0字符组成的连通块被1所完全包围。

连通的含义是，只要连续两个方块有公共边，就看做是连通。

完全包围的意思是，该连通块不与边界相接触。

Input

本题包含若干组测试数据。 每组测试数据包含： 第一行两个整数n,m表示图像的长与宽。 接下来n行m列将会是只有01组成的字符画。

满足1<=n,m<=100

Output

如果这个图是1的话，输出1；如果是0的话，输出0，都不是输出-1。

Sample Input

32 32000000000000000000000000000000000000000000011111111000000000000000000000001111111111100000000000000000000011111111111100000000000000000001111111111111100000000000000000011111100011111000000000000000001111100000011110000000000000000011111000000111110000000000000000111110000000111110000000000000011111100000001111100000000000000111111000000001111100000000000001111110000000001111000000000000011111100000000011111000000000000111110000000000111100000000000001111000000000001111000000000000011110000000000011110000000000000111100000000000011100000000000000111100000000000111000000000000001111000000000001110000000000000011110000000000011100000000000001111000000000011110000000000000011110000000000111100000000000000011100000000001111000000000000000111110000011111110000000000000001111100011111111000000000000000011111111111111100000000000000000011111111111111000000000000000001111111111111000000000000000000001111111111100000000000000000000001111111000000000000000000000000011111000000000000000000000000000000000000000000000000032 3200000000000000000000000000000000000000000000000011111100000000000000000000000000111111100000000000000000000000011111111000000000000000000000001111111110000000000000000000000001111111100000000000000000000000011111111000000000000000000000001111111100000000000000000000000011111110000000000000000000000001111111100000000000000000000000011111111100000000000000000000000111111111000000000000000000000001111111100000000000000000000000111111100000000000000000000001111111111000000000000000000001111111111111000000000000000000111111111111110000000000000000001111111111111100000000000000000011111111111110000000000000000000000011111111110000000000000000000000000011111100000000000000000000000001111111000000000000000000000001111111100000000000000000000000001111111100000000000000000000000011111111000000000000000000000000111111111000000000000000000000001111111110000000000000000000000000111111110000000000000000000000000011111111110000000000000000000000111111111100000000000000000000000111111111000000000000000000000000000000000000003 3101101011

Sample Output

Copy

01-1 dfs找连通块。1的连通块有且只有一个，0的可以不存在，如果有只存在一个，否则-1，记录下连通块的个数。 0必须被1包围，因为矩阵中只存在0和1，所以可以理解成只要不到达边界处的0即满足条件。

#include
        
         #include
         
          char a[105][105],b[105][105];int cc,ff,n,m;void dfs(int x,int y){    if(x<0||y<0||x>=n||y>=m) return;    if(a[x][y]=='0'||b[x][y]==1) return;    if(a[x][y]=='1'){        b[x][y]=1;        dfs(x+1,y);        dfs(x,y+1);        dfs(x-1,y);        dfs(x,y-1);    }}void dfss(int x,int y){    if(x<0||y<0||x>=n||y>=m){        ff=1;        return;    }    if(a[x][y]=='1'||b[x][y]==1) return;    if(a[x][y]=='0'){        b[x][y]=1;        dfss(x+1,y);        dfss(x,y+1);        dfss(x-1,y);        dfss(x,y-1);    }}int main(){    int c1,c2,co,f,i,j;    while(~scanf("%d%d",&n,&m)){        memset(a,0,sizeof(a));        memset(b,0,sizeof(b));        for(i=0;i