XQ,one of the three Sailormoon girls,is usually playing mobile games on the class.Her favorite mobile game is called “The Princess In The Wall”.Now she give you a problem about this game.
Can you solve it?The following picture show this problem better.

This game is played on a rectangular area.This area is divided into some equal square grid..There are N rows and M columns.For each grid,there may be a colored square block or nothing.
Each grid has a number.
“0” represents this grid have nothing.
“1” represents this grid have a red square block.
“2” represents this grid have a blue square block.
“3” represents this grid have a green square block.
“4” represents this grid have a yellow square block.

1. Each step,when you choose a grid have a colored square block, A group of this block and some connected blocks that are the same color would be removed from the board. no matter how many square blocks are in this group.
2. When a group of blocks is removed, the blocks above those removed ones fall down into the empty space. When an entire column of blocks is removed, all the columns to the right of that column shift to the left to fill the empty columns.

Now give you the number of the row and column and the data of each grid.You should calculate how many steps can make the entire rectangular area have no colored square blocks at least.

5 6
0 0 0 3 4 4
0 1 1 3 3 3
2 2 1 2 3 3
1 1 1 1 3 3
2 2 1 4 4 4

4

Hint

0 0 0 3 4 4     0 0 0 4 4 0    0 0 0 0 0 0    0 0 0 0 0 0    0 0 0 0 0 0
0 1 1 3 3 3     0 0 3 3 3 0    0 0 0 0 0 0    0 0 0 0 0 0    0 0 0 0 0 0
2 2 1 2 3 3     0 0 3 3 3 0    0 0 0 0 0 0    0 0 0 0 0 0    0 0 0 0 0 0
1 1 1 1 3 3     2 2 2 3 3 0    2 2 2 4 4 0    2 2 0 0 0 0    0 0 0 0 0 0
2 2 1 4 4 4     2 2 4 4 4 0    2 2 4 4 4 0    2 2 2 0 0 0    0 0 0 0 0 0
 

思路：因为方块会越消越少，所以没必要判重。注意消去之后，上面的会掉下来，如果某一列全为是空的，右边的会往左移。

#include <stdio.h>

struct{
int d[6][6],step;
}que[1000000],t;

int n,m,temp[6][6],nxt[4][2]={{0,1},{1,0},{0,-1},{-1,0}};
bool vis[6][6];

void dfs(int x,int y,int num)
{
    for(int i=0;i<4;i++)
    {
        x+=nxt[i][0];
        y+=nxt[i][1];

        if(x>=0 && x<n && y>=0 && y<m && !vis[x][y] && temp[x][y]==num)
        {
            vis[x][y]=1;

            t.d[x][y]=0;

            dfs(x,y,num);
        }

        x-=nxt[i][0];
        y-=nxt[i][1];
    }
}

int main()
{
    int i,j,k,p,q,top,bottom;
    bool flag;

    while(~scanf("%d%d",&n,&m))
    {
        for(i=0;i<n;i++) for(j=0;j<m;j++) scanf("%d",&que[0].d[i][j]);

        top=0;
        bottom=1;

        que[0].step=0;

        while(top<bottom)
        {
            t=que[top];

            flag=1;

            for(i=0;i<n && flag;i++) for(j=0;j<m && flag;j++) if(t.d[i][j]) flag=0;

            if(flag)
            {
                printf("%d\n",t.step);

                break;
            }

            t.step++;

            for(i=0;i<n;i++) for(j=0;j<m;j++) temp[i][j]=t.d[i][j],vis[i][j]=0;

            for(i=0;i<n;i++) for(j=0;j<m;j++)
            {
                if(temp[i][j] && !vis[i][j])
                {
                    vis[i][j]=1;

                    t.d[i][j]=0;

                    dfs(i,j,temp[i][j]);

                    for(p=n-1;p>=0;p--)//向下移动
                    {
                        for(q=0;q<m;q++)
                        {
                            if(!t.d[p][q])
                            {
                                for(k=p-1;k>=0;k--)
                                {
                                    if(t.d[k][q])
                                    {
                                        t.d[p][q]=t.d[k][q];
                                        t.d[k][q]=0;

                                        break;
                                    }
                                }
                            }
                        }
                    }

                    for(q=0;q<m-1;q++)//向左移动
                    {
                        for(p=0;p<n;p++) if(t.d[p][q]) break;

                        if(p<n) continue;

                        for(p=0;p<n;p++)
                        {
                            t.d[p][q]=t.d[p][q+1];
                            t.d[p][q+1]=0;
                        }
                    }

                    que[bottom++]=t;

                    for(p=0;p<n;p++) for(q=0;q<m;q++) t.d[p][q]=temp[p][q];
                }
            }

            top++;
        }
    }
}