HDU-3839-Ancient Messages(DFS)

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In order to understand early civilizations, archaeologists often study texts written in ancient languages. One such language, used in Egypt more than 3000 years ago, is based on characters called hieroglyphs. Figure C.1 shows six hieroglyphs and their names. In this problem, you will write a program to recognize these six characters.

For each test case, display its case number followed by a string containing one character for each hieroglyph recognized in the image, using the following code:

Ankh: A
Wedjat: J
Djed: D
Scarab: S
Was: W
Akhet: K

In each output string, print the codes in alphabetic order. Follow the format of the sample output.

The sample input contains descriptions of test cases shown in Figures C.2 and C.3. Due to space constraints not all of the sample input can be shown on this page.

100 25
0000000000000000000000000
0000000000000000000000000
 ...(50 lines omitted)...
00001fe0000000000007c0000
00003fe0000000000007c0000
 ...(44 lines omitted)...
0000000000000000000000000
0000000000000000000000000
150 38
00000000000000000000000000000000000000
00000000000000000000000000000000000000
    ...(75 lines omitted)...
0000000003fffffffffffffffff00000000000
0000000003fffffffffffffffff00000000000
    ...(69 lines omitted)...
00000000000000000000000000000000000000
00000000000000000000000000000000000000
0 0

Case 1: AKW
Case 2: AAAAA

思路：根据圈的数量来识别。

#include <cstdio>
#include <algorithm>
using namespace std;

char ts[201],mes[6]={'W','A','K','J','S','D'},ans[10];
bool vis[205][205];
int n,m,mp[205][205],nxt[4][2]={{1,0},{0,1},{-1,0},{0,-1}},num;

void dfs(int x,int y)
{
    int i;

    for(i=0;i<4;i++)
    {
        x+=nxt[i][0];
        y+=nxt[i][1];

        if(x>=0 && x<n && y>=0 && y<m && !vis[x][y] && !mp[x][y])
        {
            vis[x][y]=1;
            dfs(x,y);
        }

        x-=nxt[i][0];
        y-=nxt[i][1];
    }
}

void dfs3(int x,int y)
{
    int i;

    for(i=0;i<4;i++)
    {
        x+=nxt[i][0];
        y+=nxt[i][1];

        if(x>=0 && x<n && y>=0 && y<m && !vis[x][y] && !mp[x][y])
        {
            vis[x][y]=1;
            dfs3(x,y);
        }

        x-=nxt[i][0];
        y-=nxt[i][1];
    }
}

void dfs2(int x,int y)
{
    int i;

    for(i=0;i<4;i++)
    {
        x+=nxt[i][0];
        y+=nxt[i][1];

        if(x>=0 && x<n && y>=0 && y<m && !vis[x][y])
        {
            if(mp[x][y])
            {
                vis[x][y]=1;
                dfs2(x,y);
            }
            else
            {
                vis[x][y]=1;
                num++;
                dfs3(x,y);
            }

        }

        x-=nxt[i][0];
        y-=nxt[i][1];
    }
}

int main()
{
    int i,j,t,casenum=1,cnt;

    while(~scanf("%d%d",&n,&m) && n)
    {
        n++;
        m*=4;
        m++;

        for(i=0;i<=n;i++) for(j=0;j<=m;j++) vis[i][j]=0;

        for(i=1;i<n;i++)
        {
            gets(ts);

            if(!ts[0])
            {
                i--;
                continue;
            }


            for(j=0;ts[j];j++)
            {
                if(ts[j]>='a' && ts[j]<='f')
                {
                    t=ts[j]-'a'+10;

                    mp[i][j*4+1]=t/8;
                    mp[i][j*4+2]=t%8/4;
                    mp[i][j*4+3]=t%4/2;
                    mp[i][j*4+4]=t%2/1;
                }
                else
                {
                    t=ts[j]-'0';

                    mp[i][j*4+1]=t/8;
                    mp[i][j*4+2]=t%8/4;
                    mp[i][j*4+3]=t%4/2;
                    mp[i][j*4+4]=t%2/1;
                }
            }
        }

        for(i=0;i<=m;i++) mp[n][i]=0;
        for(i=0;i<=n;i++) mp[i][m]=0;

        n++;
        m++;

        vis[0][0]=1;
        dfs(0,0);

        cnt=0;

        for(i=0;i<n;i++)
        {
            for(j=0;j<m;j++)
            {
                if(mp[i][j] && !vis[i][j])
                {
                    num=0;

                    vis[i][j]=1;

                    dfs2(i,j);

                    ans[cnt++]=mes[num];
                }
            }
        }

        sort(ans,ans+cnt);

        ans[cnt]=0;

        printf("Case %d: ",casenum++);
        
        puts(ans);
    }
}

HDU-3839-Ancient Messages(DFS),布布扣,bubuko.com

HDU-3839-Ancient Messages(DFS)

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原文地址：http://blog.csdn.net/faithdmc/article/details/38365649