poj1128

maksyuki 发表于 oj 分类，标签: 图论-拓扑排序, 基础算法-搜索(DFS\BFS)\记忆化

20 3月 2016

Frame Stacking

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

.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

Source

Southern African 2001

题目类型：拓扑排序+dfs

算法分析：通过题意先将每个frame画出来，然后对于每个图画A，若有其他图画B的字符覆盖，则从A到B连一条有向边，最后dfs时优先选择字符小的即可，注意这里可能存在多个图画互相不覆盖的情况，若使用B向A建图，则会WA！！！

/*************************************************
Author        :supermaker
Created Time  :2016/3/20 10:05:27
File Location :C:\Users\abcd\Desktop\TheEternalPoet
**************************************************/

#pragma comment(linker, "/STACK:102400000,102400000")
#include <set>
#include <bitset>
#include <list>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <string>
#include <vector>
#include <ios>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <algorithm>
#include <utility>
#include <complex>
#include <numeric>
#include <functional>
#include <cmath>
#include <ctime>
#include <climits>
#include <cstdarg>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cassert>
using namespace std;

#define CFF freopen ("aaa.txt", "r", stdin)
#define CPPFF ifstream cin ("aaa.txt")
#define	DB(ccc)	cout << #ccc << " = " << ccc << endl
#define PB push_back
#define MP(A, B) make_pair(A, B)

typedef long long LL;
typedef unsigned long long ULL;
typedef double DB;
typedef pair <int, int> PII;
typedef pair <int, bool> PIB;

const int INF = 0x7F7F7F7F;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = 2 * acos (0.0);
const int maxn = 66;

int row, col, tot, ind[maxn];
char g[maxn][maxn], gg[maxn][maxn][maxn], out[maxn];
bool vis[maxn], vvis[maxn];
vector <int> edge[maxn];

void PP (char val)
{
	int minx = INF, maxx = -INF, miny = INF, maxy = -INF;
	for (int i = 1; i <= row; i++)
		for (int j = 1; j <= col; j++)
			if (g[i][j] == val)
			{
				minx = min (minx, i);
				maxx = max (maxx, i);
				miny = min (miny, j);
				maxy = max (maxy, j);
			}
	
	for (int i = minx; i <= maxx; i++)
		gg[tot][i][miny] = gg[tot][i][maxy] = val;
	for (int i = miny; i <= maxy; i++)
		gg[tot][minx][i] = gg[tot][maxx][i] = val;
	tot++;
}

void Build ()
{
	for (int i = 0; i < maxn; i++)
		edge[i].clear ();
	memset (ind, 0, sizeof (ind));
	for (int i = 0; i < tot; i++)
		for (int j = 1; j <= row; j++)
			for (int k = 1; k <= col; k++)
				if (gg[i][j][k] != '.' && gg[i][j][k] != g[j][k])
				{
					ind[g[j][k]-'A']++;
					edge[gg[i][j][k]-'A'].push_back (g[j][k] - 'A');
				}
}

void dfs (int cur)
{
	if (cur == tot)
	{
		for (int i = 0; i < tot; i++)
			printf ("%c", out[i]);
		puts ("");
		return ;
	}
	for (int i = 0; i <= 25; i++)
		if (vis[i] &&  !vvis[i] && !ind[i])
		{
			for (int j = 0; j < edge[i].size (); j++)
				ind[edge[i][j]]--;
			//printf ("%c", (char) (i + 'A'));
			out[cur] = (char) (i + 'A');
			vvis[i] = true;
			dfs (cur + 1);		
			vvis[i] = false;
			for (int j = 0; j < edge[i].size (); j++)
				ind[edge[i][j]]++;
		}
}

int main()
{
    //CFF;
    //CPPFF;
	while (scanf ("%d%d", &row, &col) != EOF)
	{
		tot = 0;
		memset (vis, false, sizeof (vis));
		for (int i = 0; i < maxn; i++)
			for (int j = 0; j < maxn; j++)
				for (int k = 0; k < maxn; k++)
					gg[i][j][k] = '.';

		for (int i = 1; i <= row; i++)
			for (int j = 1; j <= col; j++)
				scanf (" %c", &g[i][j]);
		
		for (int i = 1; i <= row; i++)
			for (int j = 1; j <= col; j++)
				if (g[i][j] != '.' && !vis[g[i][j]-'A'])
				{
					PP (g[i][j]);
					vis[g[i][j]-'A'] = true;
				}
		
		Build ();
		memset (vvis, false, sizeof (vvis));
		dfs (0);
	}
    return 0;
}

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/*************************************************

Author :supermaker

Created Time :2016/3/20 10:05:27

File Location :C:\Users\abcd\Desktop\TheEternalPoet

**************************************************/

#pragma comment(linker, "/STACK:102400000,102400000")

#include <set>

#include <bitset>

#include <list>

#include <map>

#include <stack>

#include <queue>

#include <deque>

#include <string>

#include <vector>

#include <ios>

#include <iostream>

#include <fstream>

#include <sstream>

#include <iomanip>

#include <algorithm>

#include <utility>

#include <complex>

#include <numeric>

#include <functional>

#include <cmath>

#include <ctime>

#include <climits>

#include <cstdarg>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cctype>

#include <cassert>

using namespace std;

#define CFF freopen ("aaa.txt", "r", stdin)

#define CPPFF ifstream cin ("aaa.txt")

#define DB(ccc) cout << #ccc << " = " << ccc << endl

#define PB push_back

#define MP(A, B) make_pair(A, B)

typedef long long LL;

typedef unsigned long long ULL;

typedef double DB;

typedef pair <int, int> PII;

typedef pair <int, bool> PIB;

const int INF = 0x7F7F7F7F;

const int MOD = 1e9 + 7;

const double EPS = 1e-10;

const double PI = 2 * acos (0.0);

const int maxn = 66;

int row, col, tot, ind[maxn];

char g[maxn][maxn], gg[maxn][maxn][maxn], out[maxn];

bool vis[maxn], vvis[maxn];

vector <int> edge[maxn];

void PP (char val)

{

int minx = INF, maxx = -INF, miny = INF, maxy = -INF;

for (int i = 1; i <= row; i++)

for (int j = 1; j <= col; j++)

if (g[i][j] == val)

{

minx = min (minx, i);

maxx = max (maxx, i);

miny = min (miny, j);

maxy = max (maxy, j);

}

for (int i = minx; i <= maxx; i++)

gg[tot][i][miny] = gg[tot][i][maxy] = val;

for (int i = miny; i <= maxy; i++)

gg[tot][minx][i] = gg[tot][maxx][i] = val;

tot++;

}

void Build ()

{

for (int i = 0; i < maxn; i++)

edge[i].clear ();

memset (ind, 0, sizeof (ind));

for (int i = 0; i < tot; i++)

for (int j = 1; j <= row; j++)

for (int k = 1; k <= col; k++)

if (gg[i][j][k] != '.' && gg[i][j][k] != g[j][k])

{

ind[g[j][k]-'A']++;

edge[gg[i][j][k]-'A'].push_back (g[j][k] - 'A');

}

void dfs (int cur)

{

if (cur == tot)

{

for (int i = 0; i < tot; i++)

printf ("%c", out[i]);

puts ("");

return ;

}

for (int i = 0; i <= 25; i++)

if (vis[i] && !vvis[i] && !ind[i])

{

for (int j = 0; j < edge[i].size (); j++)

ind[edge[i][j]]--;

//printf ("%c", (char) (i + 'A'));

out[cur] = (char) (i + 'A');

vvis[i] = true;

dfs (cur + 1);

vvis[i] = false;

for (int j = 0; j < edge[i].size (); j++)

ind[edge[i][j]]++;

}

int main()

{

//CFF;

//CPPFF;

while (scanf ("%d%d", &row, &col) != EOF)

{

tot = 0;

memset (vis, false, sizeof (vis));

for (int i = 0; i < maxn; i++)

for (int j = 0; j < maxn; j++)

for (int k = 0; k < maxn; k++)

gg[i][j][k] = '.';

for (int i = 1; i <= row; i++)

for (int j = 1; j <= col; j++)

scanf (" %c", &g[i][j]);

for (int i = 1; i <= row; i++)

for (int j = 1; j <= col; j++)

if (g[i][j] != '.' && !vis[g[i][j]-'A'])

{

PP (g[i][j]);

vis[g[i][j]-'A'] = true;

}

Build ();

memset (vvis, false, sizeof (vvis));

dfs (0);

}

return 0;

}

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