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
bwwbbbwbbwwbbwwwSample Output
4思路：DFS枚舉所有情況，因為每個棋子只有黑或白兩種情況，所以總共有2^16種情況代碼：#include<stdio.h>int map[5][5];int flag,step;void flip(int i,int j)//翻轉{ map[i][j]=!map[i][j]; if(i<3) map[i+1][j]=!map[i+1][j]; if(j<3) map[i][j+1]=!map[i][j+1]; if(i>0) map[i-1][j]=!map[i-1][j]; if(j>0) map[i][j-1]=!map[i][j-1];}int check()//檢測是否“清一色”{ for(int i=0; i<4; i++) for(int j=0; j<4; j++) if(map[i][j]!=map[0][0]) return 0; return 1;}void dfs(int i,int j,int dp){ if(dp==step) { flag=check(); return ; } if(flag||i==4) return ; flip(i,j);//此位置翻轉 if(j<3) dfs(i,j+1,dp+1); else dfs(i+1,0,dp+1); flip(i,j);//回溯恢復成翻轉前的狀態 if(j<3)//此位置不翻轉 dfs(i,j+1,dp); else dfs(i+1,0,dp);}int main(){ int i,j; char s; for(i=0; i<4; i++) { for(j=0; j<4; j++) { scanf("%c",&s); if(s=='b')//將圖案轉換成01圖 map[i][j]=1; else map[i][j]=0; } getchar(); } flag=0; for(step=0; step<=16; step++)//枚舉所有情況，因為總共只有16個格子，所以最多只能進行16步 { dfs(0,0,0); if(flag) break; } if(flag) printf("%d/n",step); else printf("Impossible/n"); return 0;}
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