首页 > 搜索 > DFS搜索 > hdu 3839-ancient messages-dfs-[解题报告]hoj
2015
04-13

hdu 3839-ancient messages-dfs-[解题报告]hoj

Problem Description
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.
 

Input
The input consists of several test cases, each of which describes an image containing one or more hieroglyphs chosen from among those shown in Figure C.1. The image is given in the form of a series of horizontal scan lines consisting
of black pixels (represented by 1) and white pixels (represented by 0). In the input data, each scan line is encoded in hexadecimal notation. For example, the sequence of eight pixels 10011100 (one black pixel, followed by two white pixels, and so on) would
be represented in hexadecimal notation as 9c. Only digits and lowercase letters a through f are used in the hexadecimal encoding. The first line of each test case contains two integers, H and W: H (0 < H <= 200) is the number of scan lines in the image. W
(0 < W <= 50) is the number of hexadecimal characters in each line. The next H lines contain the hexadecimal characters of the image, working from top to bottom. Input images conform to the following rules:

  • The image contains only hieroglyphs shown in Figure C.1.
  • Each image contains at least one valid hieroglyph.
  • Each black pixel in the image is part of a valid hieroglyph.
  • Each hieroglyph consists of a connected set of black pixels and each black pixel has at least one other black pixel on its top, bottom, left, or right side.
  • The hieroglyphs do not touch and no hieroglyph is inside another hieroglyph.
  • Two black pixels that touch diagonally will always have a common touching black pixel.
  • The hieroglyphs may be distorted but each has a shape that is topologically equivalent to one of the symbols in Figure C.11.

The last test case is followed by a line containing two zeros.

1Two figures are topologically equivalent if each can be transformed into the other by stretching without tearing.

 

Output
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.

 

Sample Input
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
 

Sample Output
Case 1: AKW Case 2: AAAAA
 

Source

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);
    }
}

版权声明:本文为博主原创文章,未经博主允许不得转载。


,
  1. 老实说,这种方法就是穷举,复杂度是2^n,之所以能够AC是应为题目的测试数据有问题,要么数据量很小,要么能够得到k == t,否则即使n = 30,也要很久才能得出结果,本人亲测

  2. 其实国内大部分公司对算法都不够重视。特别是中小型公司老板根本都不懂技术,也不懂什么是算法,从而也不要求程序员懂什么算法,做程序从来不考虑性能问题,只要页面能显示出来就是好程序,这是国内的现状,很无奈。