std rip puts output tex bottom ring short man

題意 給你一些矩形框堆疊後的鳥瞰圖 推斷這些矩形框的堆疊順序 每一個矩形框滿足每邊都至少有一個點可見 輸入保證至少有一個解 按字典序輸出全部可行解

和上一題有點像 僅僅是這個要打印全部的可行方案 建圖還是類似 由於每一個矩形框的四邊都有點可見 所以每一個矩形框的左上角和右下角的坐標是能夠確定的 然後一個矩形框上有其他字符時 就讓這個矩形框相應的字符和那個其他字符建立一個小於關系 由於要打印方案 所以在有多個入度為0的點時須要用DFS對每種選擇都進行一遍拓撲排序

#include <cstdio>
#include <cstring>
using namespace std;
const int N = 50;
char ans[N], g[N][N], tp[N][N];
int x1[N], y1[N], x2[N], y2[N];
//(x1,y1)為相應字母的左上角坐標  (x2,y2)為右下
int in[N], n;

void addTopo(int i, int j, int c)
{
    int t = g[i][j] - 'A';
    if(t != c && !tp[c][t])
    {
        ++in[t];
        tp[c][t] = 1;
    }
}

void build()
{
    memset(tp, 0, sizeof(tp)); //tp[i][j] = 1表示有i < j的關系
    for(int c = n = 0; c < 26; ++c)
    {
        if(in[c] < 0) continue;
        for(int i = x1[c]; i <= x2[c]; ++i)
        {
            addTopo(i, y1[c], c);
            addTopo(i, y2[c], c);
        }
        for(int j = y1[c]; j <= y2[c]; ++j)
        {
            addTopo(x1[c], j, c);
            addTopo(x2[c], j, c);
        }
        ++n;//統計出現了多少個字符
    }
}

void topoSort(int k)
{
    if(k == n)
    {
        ans[k] = 0;
        puts(ans);
        return;
    }

    //從前往後找入度為0的點保證升序
    for(int i = 0; i < 26; ++i)
    {
        if(in[i] == 0)
        {
            ans[k] = i + 'A'; //這一位放i
            in[i] = -1;
            for(int j = 0; j < 26; ++j)
                if(tp[i][j]) --in[j];

            topoSort(k + 1); //找下一位

            in[i] = 0; //回溯
            for(int j = 0; j < 26; ++j)
                if(tp[i][j]) ++in[j];
        }
    }
}

int main()
{
    int h, w, c;
    while(~scanf("%d%d", &h, &w))
    {
        for(int i = 0; i < 26; ++i)
        {
            x1[i] = y1[i] = N;
            x2[i] = y2[i] = 0;
        }

        memset(in, -1, sizeof(in));
        for(int i = 0; i < h; ++i)
        {
            scanf("%s", g[i]);
            for(int j = 0; j < w; ++j)
            {
                if((c = g[i][j] - 'A') < 0) continue; //g[i][j] ='.'
                if(i < x1[c]) x1[c] = i;
                if(i > x2[c]) x2[c] = i;
                if(j < y1[c]) y1[c] = j;
                if(j > y2[c]) y2[c] = j;
                in[c] = 0;  //出現過的字母in初始為0  否則為-1
            }
        }
        build();
        topoSort(0);
    }
    return 0;
}

Frame Stacking

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

POJ 1128 Frame Stacking（拓撲排序&#183;打印字典序）