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  • POJ 1753, Flip Game

    POJ 2965, The Pilots Brothers' refrigerator

    POJ 1753, Flip Game


    这两道题类似,可转化图论的最短路径问题:

    顶点(Vertex)数:65536

    边(Edge)数:65536 * 16 / 2

    边的权值(Weight):1

    对于给定两个点, 求最短路径:

    由于边的权值均为1, 所以适宜采用BFS + Mark Table, 时间复杂度为O(N), N为顶点数。


    Description
    Flip game is played on a rectangular 4x4 field with two-sided pieces placed on each of its 16 squares. One side of each piece is white and the other one is black and each piece is lying either it's black or white side up. Each round you flip 3 to 5 pieces, thus changing the color of their upper side from black to white and vice versa. The pieces to be flipped are chosen every round according to the following rules:
    Choose any one of the 16 pieces.
    Flip the chosen piece and also all adjacent pieces to the left, to the right, to the top, and to the bottom of the chosen piece (if there are any).

    Consider the following position as an example:

    bwbw
    wwww
    bbwb
    bwwb
    Here "b" denotes pieces lying their black side up and "w" denotes pieces lying their white side up. If we choose to flip the 1st piece from the 3rd row (this choice is shown at the picture), then the field will become:

    bwbw
    bwww
    wwwb
    wwwb
    The goal of the game is to flip either all pieces white side up or all pieces black side up. You are to write a program that will search for the minimum number of rounds needed to achieve this goal.

    Input
    The input consists of 4 lines with 4 characters "w" or "b" each that denote game field position.

    Output
    Write to the output file a single integer number - the minimum number of rounds needed to achieve the goal of the game from the given position. If the goal is initially achieved, then write 0. If it's impossible to achieve the goal, then write the word "Impossible" (without quotes).

    Sample Input
    bwwb
    bbwb
    bwwb
    bwww

    Sample Output
    4

    Source
    Northeastern Europe 2000


    // POJ1753.cpp : Defines the entry point for the console application.
    //

    #include 
    <iostream>
    #include 
    <queue>
    using namespace std;

    int minRounds(int cposition)
    {
        
    char mark[65536];
        memset(mark,
    -1,sizeof(mark));

        queue
    <int> q;
        q.push(cposition);
        mark[cposition] 
    = 0;

        
    const unsigned short pattern[16= {0xc800,0xe400,0x7200,0x3100,0x8c80,0x4e40,0x2720,0x1310,
                                            
    0x08c8,0x04e4,0x0272,0x0131,0x008c,0x004e,0x0027,0x0013 };
        
        
    int cpos = 0;
        
    int npos = 0;
        
    while (!q.empty())
        {
            cpos 
    = q.front();
            q.pop();
            
    for (int i = 0; i < 16++i) 
            {
                npos 
    = cpos ^ pattern[i];
                
    if (mark[npos] == -1)
                {
                    q.push(npos);
                    mark[npos] 
    = mark[cpos] + 1;
                }
                
    if (npos == 65535 || npos == 0return mark[npos];
            }
        }
        
    return -1;
    }


    int main(int argc, char* argv[])
    {
        
    int cposition = 0;
        
    char s[5];
        s[
    4]='\0';
        
    for(int i=1;i<=4;i++)
        {
            scanf(
    "%s",s);
            
    for(int j=0;j<=3;j++)
            {
                cposition 
    <<= 1;
                
    if(s[j]=='b')++cposition;
            }
        }

        
    int result = minRounds(cposition);
        
    if (result == -1) cout << "Impossible";
        
    else cout << result;
        
    return 0;
    }
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  • 原文地址:https://www.cnblogs.com/asuran/p/1574466.html
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