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最新推荐文章于 2020-01-22 12:03:42 发布

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最新推荐文章于 2020-01-22 12:03:42 发布

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原文链接：http://www.cnblogs.com/eric-blog/archive/2011/01/13/1934740.html

本文介绍了一款名为FlipGame的游戏算法实现，游戏目标是在最少的步骤内将所有棋子统一颜色。通过BFS和位运算的方法来寻找最优解。

Flip Game

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 12047		Accepted: 5140

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

Source

Northeastern Europe 2000

解题思路：BFS + 位运算
参考了勇泽的代码

#include<iostream>
#include<queue>
#include<cmath>

using namespace std;

typedef struct
{
    int data;
    int step;
}Node;

int Arr[16]={0xc800,0xe400,0x7200,0x3100,
             0x8c80,0x4e40,0x2720,0x1310,
             0x08c8,0x04e4,0x0272,0x0131,
             0x008c,0x004e,0x0027,0x0013};
int val;
queue<Node> Q;
bool flag[65536],exist;

void BFS()
{
    Node tmp,t;
    tmp.data=val;
    tmp.step=0;
    flag[val]=true;
    Q.push(tmp);
    if(tmp.data==0||tmp.data==65535)
    {
        cout<<0<<endl;
        exist=true;
        return ;
    }
    while(!Q.empty())
    {
        tmp=Q.front();
        Q.pop();
        for(int i=0;i<16;i++)
        {
            t.data=tmp.data^Arr[i];
            if(t.data==0||t.data==65535)
            {
                cout<<tmp.step+1<<endl;
                exist=true;
                return ;
            }
            if(!flag[t.data])
            {
                t.step=tmp.step+1;
                Q.push(t);
                flag[t.data]=true;
            }
        }
    }
}

int main()
{
    char ch;
    val=0;
    for(int i=0;i<16;i++)
    {
        cin>>ch;
        if(ch=='w')
            val+=pow(2.0,i);
    }
    BFS();
    if(!exist)
        cout<<"Impossible"<<endl;
    return 0;
}

转载于:https://www.cnblogs.com/eric-blog/archive/2011/01/13/1934740.html