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Flip Game
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:
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 |
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- Source Code
#include <iostream> using namespace std; bool m[10][10], flag; int step; bool judge() { for(int i=1;i<=4;i++) for(int j=1;j<=4;j++) if(m[i][j]!=m[1][1]) return 0; return 1; } void filp(int r, int c) { m[r][c]=!m[r][c]; m[r+1][c]=!m[r+1][c]; m[r-1][c]=!m[r-1][c]; m[r][c+1]=!m[r][c+1]; m[r][c-1]=!m[r][c-1]; } void dfs(int r, int c, int dep) { if(dep>=step){ flag=judge(); // for(int i=1;i<=4;i++){ // for(int j=1;j<=4;j++) // cout<<m[i][j]<<' '; // cout<<endl; // } // cout<<flag<<endl; return ; } if(flag||r>=5) return ; filp(r,c); if(c<4) dfs(r,c+1,dep+1); else dfs(r+1,1,dep+1); filp(r,c); if(c<4) dfs(r,c+1,dep); else dfs(r+1,1,dep); } int main() { char c; flag=0; for(int i=1;i<=4;i++) for(int j=1;j<=4;j++){ cin>>c; if(c=='b') m[i][j]=1; else m[i][j]=0; } for(step=0;step<=16;step++){ dfs(1,1,0); if(flag) break; } if(flag) cout<<step<<endl; else cout<<"Impossible"<<endl; }
本文介绍了一款名为FlipGame的游戏,该游戏在一个4x4的棋盘上进行,目标是通过最少的翻转次数使所有棋子颜色统一。文章详细解释了游戏规则,并提供了一个解决该问题的算法实现。
Consider the following position as an example:
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