UVA1003(dfs+进制转换)

题意：输入很多字符的16进制数，现在需要你先转化为2进制，之后二进制中1和0组成的字符需要你求出来。

思路：字符每一个都有所不同可以靠圆圈的数目求出。

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>

using namespace std;

int gragh[1000][1000];
int n,m;
int ans;
int dir[4][2] = {{-1,0},{0,1},{1,0},{0,-1}};

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

void dfs1(int x,int y)      //作用是把1.没用的0清掉 2.没遇到一种情况就把0清掉
{
    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){
            ans++;          //记录每一种字符的圈数
            dfs1(xx,yy);
        }
        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));

        char a[100];
        for(int i = 1;i <= n; i++){
            scanf("%s",a+1);
            for(int j = m;j >= 1; j--){
                change(a[j],i,j);       //分别是十六进制的数，第几行，第几列
            }
        }

        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;
        char s[100];
        for(int i = 1;i <= n; i++){
            for(int j = 1;j <= m; j++){
                ans = 0;
                if(gragh[i][j] == 1){
                    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;
}