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 ;

}