迭代加深搜索——PQJ2286

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The Rotation Game

Time Limit:15000MS     Memory Limit:150000KB     64bit IO Format:%I64d & %I64u

Submit  Status  Practice  POJ 2286

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

题目意思

给出24个数字(其中1,2,3各有8个) 例如下图

每一次操作 例如 A 操作把 A 旁边的数字放到 A 所在的那一行数字的另一端的末尾 然后其他数字往 A 的方向移一格

即 1 1 3 2 2 2 3 变成 1 3 2 2 2 3 1

如果执行 C 操作的话 3 2 2 2 3 1 3 就变成 3 3 2 2 2 3 1

现在问最少要多少次操作才能使中间那8个数字变成一样的数字 输出最少的操作数和中间8个数字变得一样的时候是哪一个数字

思路：迭代加深，逐层搜索，具体看代码；

#include <cstdio>
#include <cstring>
#include <queue>
#include <iostream>
using namespace std;
const int INF=1<<30;
const int MAXN=7500+10;
int maxh,c;
int num[30];
int res[10];
int A[8][7]={	1,3,7,12,16,21,23,//井字的每一笔
		2,4,9,13,18,22,24,
		11,10,9,8,7,6,5,
		20,19,18,17,16,15,14,
		24,22,18,13,9,4,2,
		23,21,16,12,7,3,1,
		14,15,16,17,18,19,20,
		5,6,7,8,9,10,11};

bool judge(int *a)//判断中间8格是否全部相同
{
	int t=a[7];
	if(a[8]!=t || a[9]!=t || a[12]!=t
		|| a[13]!=t || a[16]!=t || a[17]!=t || a[18]!=t)
	return 0;
	return 1;
}

int cal(int *a)//返回中间8格中最少的不同数字的数量
{
	int b[5]={0};
	b[a[7]]++;b[a[8]]++;b[a[9]]++;b[a[12]]++;
	b[a[13]]++;b[a[16]]++;b[a[17]]++;b[a[18]]++;
	b[0]=b[1]>b[2]?b[1]:b[2];
	b[0]=b[0]>b[3]?b[0]:b[3];
	return 8-b[0];
}

bool dfs(int deep)//逐层搜索
{
	int n=cal(num);
	if(n==0) return true;//中间8格全部相同
	if(n>maxh-deep) return false;//中间8格中最少的不同数字的数量大于还可以搜索的层数，不可能有解
	for(int i=0; i<8; i++){
		res[c++]=i;
		int tmp=num[A[i][0]];
		for(int j=1; j<7; j++){
			num[A[i][j-1]]=num[A[i][j]];
		}
		num[A[i][6]]=tmp;
		if(dfs(deep+1)) return true;
		c--;
		tmp=num[A[i][6]];
		for(int j=6; j>0; j--){
			num[A[i][j]]=num[A[i][j-1]];
		}
		num[A[i][0]]=tmp;
	}
	return false;
}

int main()
{
    //freopen("in.txt","r",stdin);
	int x;
	while(scanf("%d", &x), x)
	{
		num[1]=x;
		int i;
		for(i=2; i<=24; i++){
			scanf("%d", &x);
			num[i]=x;
		}
		if(judge(num)){
			printf("No moves needed\n");
			printf("%d\n", num[7]);
			continue;
		}
		maxh=1;
		c=0;
		while(!dfs(0))
		{
			maxh++;
		}
		for(i=0; i<c; i++){
			printf("%c",res[i]+'A');
		}
		printf("\n%d\n", num[7]);
	}
    return 0;
}