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32 32
00000000000000000000000000000000
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3 3
101
101
011

0
1
-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<stdio.h>
#include<string.h>

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<n;i++){
            getchar();
            scanf("%s",a[i]);
        }
        c1=0;c2=0;f=0;cc=0;co=0;
        for(i=0;i<n;i++){
            for(j=0;j<m;j++){
                if(a[i][j]==‘1‘&&b[i][j]==0){
                    co=1;
                    if(c1==0){
                        c1=1;
                        dfs(i,j);
                    }
                    else{
                        f=1;
                        break;
                    }
                }
                if(a[i][j]==‘0‘&&b[i][j]==0){
                    ff=0;
                    dfss(i,j);
                    if(ff==0){
                        if(c2==0) c2=1;
                        else{
                            f=1;
                            break;
                        }
                    }
                }
            }
            if(f==1) break;
        }
        if(f==1||co==0) printf("-1\n");
        else if(c2==1) printf("0\n");
        else if(c2==0) printf("1\n");
    }
    return 0;
}