提升coding能力------搜索专题(4)-----poj 2286

pku 2286 The Rotation Game
题目地址：http://acm.pku.edu.cn/JudgeOnline/problem?id=2286

题目大意：

给出一个"#"字型的图，如同行李箱上的滑轮锁一样，可以滑动这个“#”字图的行，列，有8种操作方式，问在操作次数最少的情况下，字典序最小的操作序列是多少才能使"#"字型中间的“口”的数字完全一样，IDA*。

详细分析:

待填坑。

AC代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<queue>
using namespace std;
char s[105];
int Map[9][9];
int num[5];
char path[10000000];
int cnt = 0;
int Matrix_input[8][8] = { //辅助输入的数组

   {0,0,0,0,0,0,0,0},
   {2,3,5,0,0,0,0,0},
   {2,3,5,0,0,0,0,0},
   {7,1,2,3,4,5,6,7},
   {2,3,5,0,0,0,0,0},
   {7,1,2,3,4,5,6,7},
   {2,3,5,0,0,0,0,0},
   {2,3,5,0,0,0,0,0}

};
char Matrix_rota[8][3] ={  //辅助变换的数组

  {'3','c','f'},//A 0
  {'5','c','f'},//B 1
  {'3','l','b'},//C 2
  {'5','l','b'},//D 3
  {'5','c','b'},//E 4
  {'3','c','b'},//F 5
  {'5','l','f'},//G 6
  {'3','l','f'}//H  7

};
char opt[8] ={'A','B','C','D','E','F','G','H'};
int back_opt[8]={5,4,7,6,1,0,3,2};
int h()
{
    int maxx= -1;
    memset(num,0,sizeof(num));
    for(int i = 3 ; i <= 5 ;i++ )
        for(int j= 3; j<= 5; j++)
        {
             if(!( i==4 && j== 4))
                    num[Map[i][j]]++;
             maxx = max(maxx, num[Map[i][j]]);
        }

    return 8 -maxx; //至少还要再往下搜多少层
}
void Rota(int a)
{

    int pos = Matrix_rota[a][0] -'0';
    char  type = Matrix_rota[a][1];
    char dirc = Matrix_rota[a][2];

    if(type =='l')
    {
        if(dirc == 'f')
        {
            for(int i = 1 ;i <= 7; i++)
                Map[pos][i - 1] = Map[pos][i];

            Map[pos][7] = Map[pos][0];
        }

        if(dirc == 'b')
        {
            for(int i = 7 ;i >=1; i--)
                Map[pos][i + 1] = Map[pos][i];

            Map[pos][1] = Map[pos][8];
        }
    }

    if( type == 'c' )
    {
        if( dirc == 'f' )
        {
              for(int i = 1 ;i <= 7; i++)
                Map[i - 1][pos] = Map[i][pos];

             Map[7][pos] = Map[0][pos];
        }

        if(dirc == 'b')
        {
            for(int i = 7 ;i >=1 ; i--)
                Map[i + 1][pos] = Map[i][pos];

            Map[1][pos] = Map[8][pos];
        }
    }

    return ;
}
int DFS(int deep, int n)
{


    if( h() == 0 )
        return 1;

    if(deep + h() > n)
        return 0;

    for(int i = 0 ; i< 8 ;i++)
    {
        Rota(i);

        path[ cnt++ ] = opt[i];
        if(DFS(deep + 1, n))
            return 1;

        cnt--;
        Rota(back_opt[i]);


    }

    return 0;


}
int main()
{
    while(gets(s))
    {
        cnt = 0;
        if( strlen(s) ==1 &&s[0] == '0' )
            break;

        int index = 0 ,line = 1;
        for(int i = 0 ;i < strlen(s); i++)
        {
            if(s[i]!= ' ')
            {
                index++ ;

                 if(index <= Matrix_input[line][0])
                     Map[line][Matrix_input[line][index]] = s[i] -'0';



                 if(index == Matrix_input[line][0])
                 {

                     index -= Matrix_input[line][0];

                     line++;


                 }
            }
        }




        for(int i = 1 ;  ; i++)
            if(DFS(0,i))
              break;

        if(cnt )
        {
            for(int i = 0 ; i < cnt ;i++ )
                cout << path[i];

            cout << endl;


        }

        else
            cout << "No moves needed" << endl;

         cout << Map[3][3] << endl;
        //cout << h() << endl;


        /*
        for(int i = 0 ;i < 8 ;i++)
        {
            Rota(i);

            for(int j = 1; j <=7; j++)
            {

                for(int k = 1 ;k <=7 ;k++)
                    cout << Map[j][k];

                cout << endl;
            }
            Rota(back_opt[i]);

        }
        */



    }

    return 0;

}