UVA1103

题意：输入以16进制的矩阵，先转换成2进制，之后输出形成的图案。

思路：先处理掉无关图案的0，之后一个图案一个图案的遍历，识别图案的方法就是有多少个圈圈。找到一个就全部标记为-1。并且记录圆圈的数目。

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
 
using namespace std;
 
int gragh[][];
int dir[][] = {{-,},{,},{,},{,-}};
char s[];
int n,m,ans;
 
void change(char c,int i,int j)
{
    if(c >= '' && c <= ''){
        int a = c - '';
        gragh[i][(j-)*+] = a%;a /= ;
        gragh[i][(j-)*+] = a%;a /= ;
        gragh[i][(j-)*+] = a%;a /= ;
        gragh[i][(j-)*+] = a%;
    }
    else {
        int a = c - 'a' + ;
        gragh[i][(j-)*+] = a%;a /= ;
        gragh[i][(j-)*+] = a%;a /= ;
        gragh[i][(j-)*+] = a%;a /= ;
        gragh[i][(j-)*+] = a%;
    }
}
 
void dfs1(int x,int y)
{
    gragh[x][y] = -;
    for(int i = ;i < ; i++){
        int xx = x + dir[i][];
        int yy = y + dir[i][];
        if(xx >=  && xx <= n && yy >=  && yy <= m && gragh[xx][yy] == ){
            dfs1(xx,yy);
        }
    }
}
 
void dfs2(int x,int y)
{
    gragh[x][y] = -;
    for(int i = ;i < ; i++){
        int xx = x + dir[i][];
        int yy = y + dir[i][];
        if(xx >=  && xx <= n && yy >=  && yy <= m && gragh[xx][yy] == ){
            dfs1(xx,yy);
            ans++;
        }
        else if(xx >=  && xx <= n && yy >=  && yy <= m && gragh[xx][yy] == ){
            dfs2(xx,yy);
        }
    }
}
 
int main()
{
  //  freopen("in.txt","r",stdin);
    int ncase = ;
    while(scanf("%d%d",&n,&m) != EOF){
        if(n ==  && m == )
            break;
 
        memset(gragh,,sizeof(gragh));
        memset(s,,sizeof(s));
        char c[];
        for(int i = ;i <= n; i++){
            scanf("%s",c+);
            for(int j = m;j >= ; j--){
                change(c[j],i,j);
            }
        }
        m *=;
 
        for(int i = ;i <= n; i++){
            if(gragh[i][] == )
                dfs1(i,);
            if(gragh[i][m] == )
                dfs1(i,m);
        }
        for(int j = ;j <= m; j++){
            if(gragh[][j] == )
                dfs1(,j);
            if(gragh[n][j] == )
                dfs1(n,j);
        }
        int pos = ;
        for(int i = ;i <= n; i++){
            for(int j = ;j <= m; j++){
                if(gragh[i][j] == ){
                    ans = ;
                    dfs2(i,j);
                    if(ans == ){
                        s[pos++] = 'A';
                    }
                    else if(ans == )
                        s[pos++] = 'J';
                    else if(ans == )
                        s[pos++] = 'D';
                    else if(ans == )
                        s[pos++] = 'S';
                    else if(ans == )
                        s[pos++] = 'W';
                    else if(ans == )
                        s[pos++] = 'K';
                }
            }
        }
 
        sort(s,s+pos);
        printf("Case %d: ",ncase++);
        for(int i = ;i < pos; i++)
            printf("%c",s[i]);
        printf("\n");
    }
    return ;
}