【拓扑排序+dfs】Frame Stacking POJ - 1128

Frame Stacking POJ - 1128 

Consider the following 5 picture frames placed on an 9 x 8 array. 

........ ........ ........ ........ .CCC....

EEEEEE.. ........ ........ ..BBBB.. .C.C....

E....E.. DDDDDD.. ........ ..B..B.. .C.C....

E....E.. D....D.. ........ ..B..B.. .CCC....

E....E.. D....D.. ....AAAA ..B..B.. ........

E....E.. D....D.. ....A..A ..BBBB.. ........

E....E.. DDDDDD.. ....A..A ........ ........

E....E.. ........ ....AAAA ........ ........

EEEEEE.. ........ ........ ........ ........

    1        2        3        4        5   


Now place them on top of one another starting with 1 at the bottom and ending up with 5 on top. If any part of a frame covers another it hides that part of the frame below. 

Viewing the stack of 5 frames we see the following. 

.CCC....

ECBCBB..

DCBCDB..

DCCC.B..

D.B.ABAA

D.BBBB.A

DDDDAD.A

E...AAAA

EEEEEE..






In what order are the frames stacked from bottom to top? The answer is EDABC. 

Your problem is to determine the order in which the frames are stacked from bottom to top given a picture of the stacked frames. Here are the rules: 

1. The width of the frame is always exactly 1 character and the sides are never shorter than 3 characters. 

2. It is possible to see at least one part of each of the four sides of a frame. A corner shows two sides. 

3. The frames will be lettered with capital letters, and no two frames will be assigned the same letter.

Input

Each input block contains the height, h (h<=30) on the first line and the width w (w<=30) on the second. A picture of the stacked frames is then given as h strings with w characters each. 
Your input may contain multiple blocks of the format described above, without any blank lines in between. All blocks in the input must be processed sequentially.

Output

Write the solution to the standard output. Give the letters of the frames in the order they were stacked from bottom to top. If there are multiple possibilities for an ordering, list all such possibilities in alphabetical order, each one on a separate line. There will always be at least one legal ordering for each input block. List the output for all blocks in the input sequentially, without any blank lines (not even between blocks).

Sample Input

9
8
.CCC....
ECBCBB..
DCBCDB..
DCCC.B..
D.B.ABAA
D.BBBB.A
DDDDAD.A
E...AAAA
EEEEEE..

Sample Output

EDABC

题意  

给出叠加后的效果图,去判断是按照什么顺序叠加起来的,

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规则如下:

(1)    图片宽度均为一个字符,四边均不短于三个字符;

(2)    每张图片四条边当中每条边都能看见一部分;

(3)    每张图片中字母代表对应的图片,任何两张图片中不会有相同的字母。

 由于要打印方案  所以在有多个入度为0的点时需要用DFS对每种选择都进行一遍拓扑排序

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
char a[35][35];
int ans[30];
int G[30][30];
struct node
{
    int sx,sy,ex,ey;
} b[30];
int vis[30];
int du[30];
int tot;
int n,m;

void init()
{
    memset(du,-1,sizeof(du));
    memset(vis,0,sizeof vis);
    tot=0;
    for(int i=0; i<26; i++)
    {
        b[i].sx=100;
        b[i].ex=-1;
        b[i].sy=100;
        b[i].ey=-1;
    }
    memset(G,0,sizeof G);
    memset(ans,0,sizeof ans);
}

void dfs(int num)
{
    if(num==tot)
    {
        for(int i=0; i<tot; i++)
        {
            char k=ans[i]+'A';
            cout<<k;
        }
        cout<<endl;
        return ;
    }

    for(int i=0; i<26; i++)
    {
        if(du[i]==0)
        {
            du[i]=-1;
            ans[num]=i;
            for(int j=0; j<26; j++)
            {
                if(G[i][j])
                    du[j]--;
            }
            dfs(num+1);

            du[i]=0;
            for(int j=0; j<26; j++)
            {
                if(G[i][j])
                    du[j]++;
            }
        }
    }
}

int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        init();
        for(int i=1; i<=n; i++)
            scanf("%s",a[i]+1);
        for(int i=1; i<=n; i++)
        {
            for(int j=1; j<=m; j++)
            {
                if(a[i][j]=='.') continue;
                int k=a[i][j]-'A';
                if(!vis[k])
                {
                    du[k]=0;
                    vis[k]=1;
                    tot++;
                }
                b[k].sx=min(b[k].sx,i);
                b[k].ex=max(b[k].ex,i);
                b[k].sy=min(b[k].sy,j);
                b[k].ey=max(b[k].ey,j);
            }
        }
        for(int k=0; k<26; k++)
        {
            if(vis[k])
            {
                for(int i=b[k].sx; i<=b[k].ex; i++)
                {
                    int u=a[i][b[k].sy]-'A';
                    int v=a[i][b[k].ey]-'A';
                    if(!G[k][u]&&k!=u)
                    {
                        G[k][u]=1;
                        du[u]++;
                    }
                    if(!G[k][v]&&k!=v)
                    {
                        G[k][v]=1;
                        du[v]++;
                    }
                }

                for(int i=b[k].sy; i<=b[k].ey; i++)
                {
                    int u=a[b[k].sx][i]-'A';
                    int v=a[b[k].ex][i]-'A';
                    if(!G[k][u]&&k!=u)
                    {
                        G[k][u]=1;
                        du[u]++;
                    }
                    if(!G[k][v]&&k!=v)
                    {
                        G[k][v]=1;
                        du[v]++;
                    }
                }
            }
        }

        dfs(0);
    }


    return 0;
}

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转载自blog.csdn.net/qq_41037114/article/details/89314515