【IDA*】——poj2286——井格移棋

                                        The Rotation Game

Time Limit: 15000MS		Memory Limit: 150000K
Total Submissions: 6021		Accepted: 2023

Description

The rotation game uses a # shaped board, which can hold 24 pieces of square blocks (see Fig.1). The blocks are marked with symbols 1, 2 and 3, with exactly 8 pieces of each kind. 

Initially, the blocks are placed on the board randomly. Your task is to move the blocks so that the eight blocks placed in the center square have the same symbol marked. There is only one type of valid move, which is to rotate one of the four lines, each consisting of seven blocks. That is, six blocks in the line are moved towards the head by one block and the head block is moved to the end of the line. The eight possible moves are marked with capital letters A to H. Figure 1 illustrates two consecutive moves, move A and move C from some initial configuration. 

Input

The input consists of no more than 30 test cases. Each test case has only one line that contains 24 numbers, which are the symbols of the blocks in the initial configuration. The rows of blocks are listed from top to bottom. For each row the blocks are listed from left to right. The numbers are separated by spaces. For example, the first test case in the sample input corresponds to the initial configuration in Fig.1. There are no blank lines between cases. There is a line containing a single `0' after the last test case that ends the input. 

Output

For each test case, you must output two lines. The first line contains all the moves needed to reach the final configuration. Each move is a letter, ranging from `A' to `H', and there should not be any spaces between the letters in the line. If no moves are needed, output `No moves needed' instead. In the second line, you must output the symbol of the blocks in the center square after these moves. If there are several possible solutions, you must output the one that uses the least number of moves. If there is still more than one possible solution, you must output the solution that is smallest in dictionary order for the letters of the moves. There is no need to output blank lines between cases. 

Sample Input

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

Sample Output

AC

2

DDHH

2

题意：有八种操作棋盘进行移动，使得中间8个数字一样，问·最短移动步数及如何移动

代码如下：

#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<stdlib.h>
using namespace std;
int a[25];
int p[8] = {6,7,8,11,12,15,16,17};//中间8个所对应的序号
int rev[8] = {5,4,7,6,1,0,3,2};//A-F B-E...反着移动
int v,ans[110];
int po[8][7] = {0,2,6,11,15,20,22,
            1,3,8,12,17,21,23,
            10,9,8,7,6,5,4,
            19,18,17,16,15,14,13,
            23,21,17,12,8,3,1,
            22,20,15,11,6,2,0,
            13,14,15,16,17,18,19,
            4,5,6,7,8,9,10};//8种操作的原始顺序 对应ABCDEFGH
void change(int k)//操作一次的结果
{
    int i,y = a[po[k][0]];
    for(i = 0 ; i < 6 ; i++)
    a[po[k][i]] = a[po[k][i+1]];
    a[po[k][6]] = y;
}
int fdep()//这个是简单的估计下还需要搜得层数 假如中间已经有5个相同的了 那最少还要移3次
{
    int i,x[4] = {0,0,0,0};
    for(i = 0 ; i < 8 ; i++)
    x[a[p[i]]]++;
    int an=0;
    for(i = 1 ; i < 4 ; i++)
    an = max(an,x[i]);
    return 8-an;
}
int dfs(int depth)
{
    int i,tt;
    for(i = 0 ; i < 8 ; i++)
    {
        change(i);//操作
        tt = fdep();
        if(tt==0)//已经到达目的解
        {
            ans[depth] = i;
            return 1;
        }
        if(depth+tt<v)//如果没有超过层数限制
        {
            ans[depth] = i;
            if(dfs(depth+1))
            return 1;
        }
        change(rev[i]);//撤销操作
    }
    return 0;
}
int main()
{
    int i;
    while(scanf("%d",&a[0])&&a[0])
    {
        for(i = 1 ; i < 24 ; i++)
        scanf("%d",&a[i]);
        if(fdep()==0)
        {
            puts("No moves needed");
            printf("%d\n",a[17]);//这里不要忘了输出
            continue;
        }
        v = 1;
        while(!dfs(0))
        {
            v++;
        }
        for(i = 0 ; i < v ; i++)
        printf("%c",ans[i]+'A');
        puts("");
        printf("%d\n",a[15]);
    }
    return 0;
}

 

 

 

优化：

两个剪枝：

a)当前操作是上一次操作的逆操作。（此题并未优化许多）

b)当前状态最好情况也无法在depth之内完成任务，即使中间8个格子中最多的数字在最好情况下凑成目标态也超过了depth。