POJ - 1128 Frame Stacking （拓扑排序+dfs）

最新推荐文章于 2018-07-23 15:44:46 发布

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最新推荐文章于 2018-07-23 15:44:46 发布

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该博客详细解析了POJ 1128题目的解决方案，通过拓扑排序和深度优先搜索（DFS）算法，确定由大写字母组成的方框从下到上的所有可能排列顺序。重点在于理解如何处理入度为0的框，并反向操作拓扑排序以满足题目中从下到上的字典序要求。

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题目链接：http://poj.org/problem?id=1128点击打开链接

Frame Stacking

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 5646		Accepted: 1973

Description

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

题目的大概意思是有一堆由相同大写字母组成的方框然后给你从上往下看的图让你求出所有的可能的从下到上的序列

很明显从上往下看的时候如果某框是完整的那么说明他的入度为0

其他的只需要遍历框就可以记录每个框上面还有几个不同的框相应的入度为多少

但是这题是从下到上因此对于一般的拓扑进行反向操作。。

并且因为要求字典序因此从A开始寻找

如此一来便有所有序列

代码有点长

#include <stdio.h>
#include <set>
#include <vector>
#include <string>
#include <iostream>
#include <algorithm>
#include <limits.h>
using namespace std;
char mmap[33][33];
int n,m;
struct xjy
{
	int flag;
	int x1;
	int x2;
	int y1;
	int y2;
};
xjy re[30];
set<int >s[30];
int rudu[30];
set<int >::iterator it;
int cnt;
int flag[30];
void dfs(string str,int num)
{
	if(num==cnt)
	{
		cout << str << endl;
		return ;
	}
	for(int i=0;i<n;i++)
	{
		if(rudu[i]==0&&flag[i]==0)
		{
			flag[i]=1;
			vector<int > mmid;
			for(it=s[i].begin();it!=s[i].end();it++)
			{
				mmid.push_back(*it);
				rudu[*it]--;
			}
			dfs(str+char('A'+i),num+1);
			flag[i]=0;
			for(int j=0;j<mmid.size();j++)
				rudu[mmid[j]]++;
		}
	}
}
int main()
{
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		for(int i=0;i<33;i++)
			for(int j=0;j<33;j++)
				mmap[i][j]='.';
		for(int i=0;i<30;i++)
		{
			rudu[i]=1;
			s[i].clear();
			flag[i]=0;
		}
		cnt=0;
		for(int i=0;i<30;i++)
		{
			re[i].flag=0;
			re[i].x1=INT_MAX;
			re[i].x2=INT_MIN;
			re[i].y1=INT_MAX;
			re[i].y2=INT_MIN;
		}

		for(int i=0;i<n;i++)
			for(int j=0;j<m;j++)
			{
				scanf(" %c",&mmap[i][j]);
				if(mmap[i][j]!='.')
				{
					int num=mmap[i][j]-'A';
					re[num].flag=1;
					rudu[num]=0;
					re[num].x1=min(re[num].x1,i);
					re[num].x2=max(re[num].x2,i);
					re[num].y1=min(re[num].y1,j);
					re[num].y2=max(re[num].y2,j);
				}
			}
		for(int i=0;i<n;i++)
		{
			if(re[i].flag)
			{
				cnt++;
				for(int j=re[i].x1;j<=re[i].x2;j++)
				{
					if(mmap[j][re[i].y1]!=char('A'+i))
					{
						if(!s[i].count(mmap[j][re[i].y1]-'A'))
						{

							rudu[mmap[j][re[i].y1]-'A']++;
							s[i].insert(mmap[j][re[i].y1]-'A');
						}
					}

					if(mmap[j][re[i].y2]!=char('A'+i))
					{

						if(!s[i].count(mmap[j][re[i].y2]-'A'))
						{
							rudu[mmap[j][re[i].y2]-'A']++;
							s[i].insert(mmap[j][re[i].y2]-'A');
						}
					}

				}
				for(int j=re[i].y1;j<=re[i].y2;j++)
				{

					if(mmap[re[i].x1][j]!=char('A'+i))
					{

						if(!s[i].count(mmap[re[i].x1][j]-'A'))
						{
							rudu[mmap[re[i].x1][j]-'A']++;
							s[i].insert(mmap[re[i].x1][j]-'A');
						}
					}
					if(mmap[re[i].x2][j]!=char('A'+i))
					{

						if(!s[i].count(mmap[re[i].x2][j]-'A'))
						{
							rudu[mmap[re[i].x2][j]-'A']++;
							s[i].insert(mmap[re[i].x2][j]-'A');
						}
					}
				}
			}
		}
		dfs("",0);
	}
}