POJ1128 Frame Stacking（拓扑排序）经典

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.

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.

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).

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

EDABC

Southern African 2001

题意：一个二维图里面有几个相框（四条边的空心矩形框，边框由同一个字符围成的），每个边框字符都不相同。边框有重叠，求重叠顺序。

1.多组输出,有不确定的情况全部输出,按字典顺序排列.
2.图中的的frame均会给出4条边的情况(一个顶点包括2条边),可以推断出frame的长度和位置.

思路：
1.矩形的判定，由条件可知，每个矩形可以用四条边表示，横坐标最小upx与最大dowx，纵坐标最小ly与最大ry，来唯一确定。然后遍历这四条边确定出来的边框A。找出哪些边框B凌驾于该边框之上，则添加一条有向边从A到B。
2.然后就是DFS式的拓扑排序。
#include<stdio.h>
#include<string.h>
#include<vector>
using namespace std;
const int N = 40;
struct frame
{
    int ux,dx,ly,ry;  //框架的四条边
};
bool exist[N]; 
int in[N],n,path[N];  
vector<int>mapt[N];

void tope(int u,int k)
{
    path[k]=u;
    if(k==n)
    {
        for(int i=1;i<=k;i++)
            printf("%c",path[i]+'A');
        printf("\n");
        return ;
    }

    in[u]=-1;
    int len=mapt[u].size();
    for(int i=0;i<len;i++)//去边
        in[mapt[u][i]]--;

    for(int i=0;i<26;i++)
     if(exist[i]&&in[i]==0)//条件：有i类边框，并且入度为0
     tope(i,k+1);

    for(int i=0;i<len;i++)//还原边
        in[mapt[u][i]]++;
    in[u]=0;
}

char str[N][N];

void buildMap(frame b[])
{
    bool have[N][N]={0};
    int ch,l,r,x,y;

    for(int i=0;i<26;i++)
    if(exist[i])
    {
        x=b[i].ux;
        l=b[i].ly; r=b[i].ry;
        while(l<=r)
        {
            if(str[x][l]!='.'&&str[x][l]-'A'!=i)
            { ch=str[x][l]-'A'; have[i][ch]=1; }
            l++;
        }
        x=b[i].dx;
        l=b[i].ly; r=b[i].ry;
        while(l<=r)
        {
            if(str[x][l]!='.'&&str[x][l]-'A'!=i)
            { ch=str[x][l]-'A'; have[i][ch]=1; }
            l++;
        }
        y=b[i].ly;
        l=b[i].ux; r=b[i].dx;
        while(l<=r)
        {
            if(str[l][y]!='.'&&str[l][y]-'A'!=i)
            { ch=str[l][y]-'A'; have[i][ch]=1; }
            l++;
        }
        y=b[i].ry;
        l=b[i].ux; r=b[i].dx;
        while(l<=r)
        {
            if(str[l][y]!='.'&&str[l][y]-'A'!=i)
            { ch=str[l][y]-'A'; have[i][ch]=1; }
            l++;
        }

      for(int j=0;j<26;j++)
      if(have[i][j])
      {
          in[j]++; mapt[i].push_back(j);  //j类边框重在i类边框之上
      }
    }
}
int main()
{
    frame board[N];
    int h,w;

    while(scanf("%d",&h)>0)
    {
        scanf("%d",&w);

        n=0;
        for(int i=0;i<=30;i++)
        {
            mapt[i].clear();
            in[i]=0;
            exist[i]=false;
            board[i].ux=board[i].ly=N;
            board[i].dx=board[i].ry=-1;
        }

        for(int i=0;i<h;i++)
        {
            scanf("%s",str[i]);
            for(int j=0;j<w;j++)
            if(str[i][j]!='.')
            {
                int ch=str[i][j]-'A';
                exist[ch]=true;
                if(board[ch].ux>i) board[ch].ux=i;
                if(board[ch].dx<i) board[ch].dx=i;
                if(board[ch].ly>j) board[ch].ly=j;
                if(board[ch].ry<j) board[ch].ry=j;
            }
        }

        buildMap(board);
        for(int i=0;i<26;i++)
        if(exist[i])
            n++;  //记录有多少个不同类型的边框
        for(int i=0;i<26;i++)
        if(exist[i]&&in[i]==0)
            tope(i,1);
    }
}