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[1000][1000];
int dir[8][2] = {{-1,0},{0,1},{1,0},{0,-1}};
char s[100];
int n,m,ans;

void change(char c,int i,int j)
{
    if(c >= '0' && c <= '9'){
        int a = c - '0';
        gragh[i][(j-1)*4+4] = a%2;a /= 2;
        gragh[i][(j-1)*4+3] = a%2;a /= 2;
        gragh[i][(j-1)*4+2] = a%2;a /= 2;
        gragh[i][(j-1)*4+1] = a%2;
    }
    else {
        int a = c - 'a' + 10;
        gragh[i][(j-1)*4+4] = a%2;a /= 2;
        gragh[i][(j-1)*4+3] = a%2;a /= 2;
        gragh[i][(j-1)*4+2] = a%2;a /= 2;
        gragh[i][(j-1)*4+1] = a%2;
    }
}

void dfs1(int x,int y)
{
    gragh[x][y] = -1;
    for(int i = 0;i < 4; i++){
        int xx = x + dir[i][0];
        int yy = y + dir[i][1];
        if(xx >= 1 && xx <= n && yy >= 1 && yy <= m && gragh[xx][yy] == 0){
            dfs1(xx,yy);
        }
    }
}

void dfs2(int x,int y)
{
    gragh[x][y] = -1;
    for(int i = 0;i < 4; i++){
        int xx = x + dir[i][0];
        int yy = y + dir[i][1];
        if(xx >= 1 && xx <= n && yy >= 1 && yy <= m && gragh[xx][yy] == 0){
            dfs1(xx,yy);
            ans++;
        }
        else if(xx >= 1 && xx <= n && yy >= 1 && yy <= m && gragh[xx][yy] == 1){
            dfs2(xx,yy);
        }
    }
}

int main()
{
  //  freopen("in.txt","r",stdin);
    int ncase = 1;
    while(scanf("%d%d",&n,&m) != EOF){
        if(n == 0 && m == 0)
            break;

        memset(gragh,0,sizeof(gragh));
        memset(s,0,sizeof(s));
        char c[1000];
        for(int i = 1;i <= n; i++){
            scanf("%s",c+1);
            for(int j = m;j >= 1; j--){
                change(c[j],i,j);
            }
        }
        m *=4;

        for(int i = 1;i <= n; i++){
            if(gragh[i][1] == 0)
                dfs1(i,1);
            if(gragh[i][m] == 0)
                dfs1(i,m);
        }
        for(int j = 1;j <= m; j++){
            if(gragh[1][j] == 0)
                dfs1(1,j);
            if(gragh[n][j] == 0)
                dfs1(n,j);
        }
        int pos = 0;
        for(int i = 1;i <= n; i++){
            for(int j = 1;j <= m; j++){
                if(gragh[i][j] == 1){
                    ans = 0;
                    dfs2(i,j);
                    if(ans == 1){
                        s[pos++] = 'A';
                    }
                    else if(ans == 3)
                        s[pos++] = 'J';
                    else if(ans == 5)
                        s[pos++] = 'D';
                    else if(ans == 4)
                        s[pos++] = 'S';
                    else if(ans == 0)
                        s[pos++] = 'W';
                    else if(ans == 2)
                        s[pos++] = 'K';
                }
            }
        }

        sort(s,s+pos);
        printf("Case %d: ",ncase++);
        for(int i = 0;i < pos; i++)
            printf("%c",s[i]);
        printf("n");
    }
    return 0;
}