POJ 1753, Flip Game

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最新推荐文章于 2021-03-14 22:03:04 发布

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原文链接：http://www.cnblogs.com/asuran/archive/2009/09/26/1574466.html

本文介绍如何将FlipGame转化为图论最短路径问题，使用BFS算法找到所有棋子颜色统一所需的最少翻转次数。针对4x4棋盘，通过定义顶点和边，实现状态转移，解决游戏目标。

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 == 0) return 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;
}

转载于:https://www.cnblogs.com/asuran/archive/2009/09/26/1574466.html