POJ-2286-The Rotation Game(IDA*入门)

本文介绍了一个名为“旋转游戏”的算法问题,目标是通过旋转特定的行来使中间区域的方块符号一致。文章详细解释了游戏规则、输入输出格式及样例,并提供了一种结合A*算法与迭代深化搜索(IDA*)的解决方案。

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The Rotation Game
Time Limit: 15000MS Memory Limit: 150000K
Total Submissions: 6299 Accepted: 2121
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.
这里写图片描述

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' toH’, 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

思路如下

这里写图片描述

A*+迭代深搜=IDA*

代码

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;
int num[25];
string str;
void A()
{
    int flag=num[1];
    num[1]=num[3];
    num[3]=num[7];
    num[7]=num[12];
    num[12]=num[16];
    num[16]=num[21];
    num[21]=num[23];
    num[23]=flag;
}
void B()
{
    int flag=num[2];
    num[2]=num[4];
    num[4]=num[9];
    num[9]=num[13];
    num[13]=num[18];
    num[18]=num[22];
    num[22]=num[24];
    num[24]=flag;
}
void C()
{
    int flag=num[11];
    for(int i=11; i>=6; i--)
        num[i]=num[i-1];
    num[5]=flag;
}
void D()
{
    int flag=num[20];
    for(int i=20; i>=15; i--)
        num[i]=num[i-1];
    num[14]=flag;
}
void E()
{
    int flag=num[24];
    num[24]=num[22];
    num[22]=num[18];
    num[18]=num[13];
    num[13]=num[9];
    num[9]=num[4];
    num[4]=num[2];
    num[2]=flag;
}
void F()
{
    int flag=num[23];
    num[23]=num[21];
    num[21]=num[16];
    num[16]=num[12];
    num[12]=num[7];
    num[7]=num[3];
    num[3]=num[1];
    num[1]=flag;
}
void G()
{
    int flag=num[14];
    for(int i=14; i<=19; i++)
        num[i]=num[i+1];
    num[20]=flag;
}
void H()
{
    int flag=num[5];
    for(int i=5; i<=10; i++)
        num[i]=num[i+1];
    num[11]=flag;
}
int get_H()
{
    int AC[4]= {0};
    AC[num[7]]++;
    AC[num[8]]++;
    AC[num[9]]++;
    AC[num[12]]++;
    AC[num[13]]++;
    AC[num[16]]++;
    AC[num[17]]++;
    AC[num[18]]++;
    return 8-max(AC[1],max(AC[2],AC[3]));
}
bool DFS(int steap,int max_steap)
{
    if(get_H()==0)
    {
        if((int)str.size()==0)
            cout<<"No moves needed"<<endl<<num[7]<<endl;
        else
            cout<<str<<endl<<num[7]<<endl;
        return true;
    }
    if(steap+get_H()>max_steap)
        return false;
    A();
    str+="A";
    if(DFS(steap+1,max_steap)==true)
        return true;
    str.erase(str.end()-1,str.end());
    F();

    B();
    str+="B";
    if(DFS(steap+1,max_steap)==true)
        return true;
    str.erase(str.end()-1,str.end());
    E();

    C();
    str+="C";
    if(DFS(steap+1,max_steap)==true)
        return true;
    str.erase(str.end()-1,str.end());
    H();

    D();
    str+="D";
    if(DFS(steap+1,max_steap)==true)
        return true;
    str.erase(str.end()-1,str.end());
    G();

    E();
    str+="E";
    if(DFS(steap+1,max_steap)==true)
        return true;
    str.erase(str.end()-1,str.end());
    B();

    F();
    str+="F";
    if(DFS(steap+1,max_steap)==true)
        return true;
    str.erase(str.end()-1,str.end());
    A();

    G();
    str+="G";
    if(DFS(steap+1,max_steap)==true)
        return true;
    str.erase(str.end()-1,str.end());
    D();

    H();
    str+="H";
    if(DFS(steap+1,max_steap)==true)
        return true;
    str.erase(str.end()-1,str.end());
    C();
    return false;
}
int main()
{
    while(scanf("%d",&num[1])&&num[1])
    {
        for(int i=2; i<=24; i++)
            scanf("%d",&num[i]);
        int result=0;
        str.clear();
        while(DFS(0,result)==false)
            result++,str.clear();
    }
    return 0;
}
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