hdu 3295 模拟过程。数据很水

An interesting mobile game

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 255    Accepted Submission(s): 123

Problem Description

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.

 

Input

There are multiple test cases. Each case starts with two positive integer N, M,(N, M <= 6)the size of rectangular area. Then n lines follow, each contains m positive integers X.(0<= X <= 4)It means this grid have a colored square block or nothing.

 

Output

Please output the minimum steps.

 

Sample Input

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

Sample Output

4

Hint

Author

B.A.C

 

Source

2010 “HDU-Sailormoon” Programming Contest

 

前几天做的题目，现在题意就忘记了，人才啊。

题意：一个游戏，问消除的最小步数是多少。只要4个方向有相同的就可以一起消除。

        如果下方有空的，就覆盖掉空的。如果前面一列是空的，则覆盖那一列。

        表达不是很好，给自己看的。

#include<iostream>
#include<stdio.h>
#include<cstring>
#include<cstdlib>
#include<queue>
using namespace std;

struct node
{
    int a[7][7];
    int step;
}start;
struct st
{
    int x,y;
};
int n,m;
int to[4][2]={ {1,0},{0,1},{0,-1},{-1,0} };
bool use[7][7];
bool hash[7][7];
queue<node>Q;
queue<st>S;

void Init_use(node &t)
{
    int i,j;
    for(i=1;i<=n;i++)
        for(j=1;j<=m;j++)
            if(t.a[i][j]==0)use[i][j]=true;
            else use[i][j]=false;
}
bool pd(node &t)
{
    int i,j;
    bool flag=true;
    for(i=1;i<=n;i++)
        for(j=1;j<=m;j++)
            if(t.a[i][j]!=0)flag=false;
    return flag;
}
void change(node &t)
{
    int i,j;
    int tmp[7][7],k,s;
    bool flag;

    for(i=1;i<=n;i++)
    {
        for(j=1;j<=m;j++)
            if(!use[i][j] && hash[i][j])
            {
                use[i][j]=true;
                t.a[i][j]=0;
            }
    }
    memset(tmp,0,sizeof(tmp));
    for(i=1;i<=m;i++)
    {
        k=n;
        for(j=n;j>=1;j--)
            if(t.a[j][i]>0)
                tmp[k--][i]=t.a[j][i];
    }
    memset(t.a,0,sizeof(t.a));
    for(i=1,s=1;i<=m;i++)
    {
        flag=true;
        for(j=1;j<=n;j++) if(tmp[j][i]!=0){flag=false;break;}
        if(flag)continue;
        for(j=1;j<=n;j++)
            t.a[j][s]=tmp[j][i];
        s++;
    }
}
void bfs(int x,int y,node t,int num)
{
    int i;
    st ans,cur;
    memset(hash,false,sizeof(hash));
    while(!S.empty())
    {
        S.pop();
    }
    ans.x=x;
    ans.y=y;
    S.push(ans);
    hash[x][y]=true;

    while(!S.empty())
    {
        cur=S.front();
        S.pop();

        for(i=0;i<4;i++)
        {
            ans=cur;
            ans.x=ans.x+to[i][0];
            ans.y=ans.y+to[i][1];
            if(ans.x>=1&&ans.x<=n && ans.y>=1&&ans.y<=m && !hash[ans.x][ans.y])
            {
                if(t.a[ans.x][ans.y]==num)
                {
                    hash[ans.x][ans.y]=true;
                    S.push(ans);
                }
            }
        }
    }
}
void dbfs()
{
    int i,j;
    node cur,t;
    Q.push(start);
    while(!Q.empty())
    {
        cur=Q.front();
        Q.pop();

        if(pd(cur)==true)
        {
            printf("%d
",cur.step);
            return;
        }
        Init_use(cur);

        for(i=1;i<=n;i++)
            for(j=1;j<=m;j++)
            {
                if(use[i][j]==false && cur.a[i][j]!=0)
                {
                    t=cur;
                    bfs(i,j,t,cur.a[i][j]);
                    change(t);
                    t.step++;
                    Q.push(t);
                }
            }
    }
}
int main()
{
    int i,j;
    while(scanf("%d%d",&n,&m)>0)
    {
        for(i=1;i<=n;i++)
            for(j=1;j<=m;j++)
                scanf("%d",&start.a[i][j]);
        start.step=0;
        while(!Q.empty())
        {
            Q.pop();
        }
        dbfs();
    }
    return 0;
}