poj 1753 Flip Game (dfs)

Flip Game

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 28805		Accepted: 12461

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

 1 //164K    125MS    C++    1211B    2014-04-26 11:02:12
 2 /*
 3 
 4     题意：
 5         问最少翻几步可以使棋盘棋子一样，不可能就输出Impossible 
 6 
 7     搜索枚举：
 8         枚举全部状态，每个位置的棋子有翻或不翻两种状态，枚举全部状态。
 9     注意一个棋子翻两次则和没翻一样，所以一种有2^16种情况，用dfs枚举全部状态。 
10     
11 */
12 #include<stdio.h>
13 #include<string.h>
14 int g[10][10];
15 int flag;
16 int judge(int tg[][10]) //判断 
17 {
18     for(int i=1;i<=4;i++)
19         for(int j=1;j<=4;j++)
20             if(g[i][j]!=g[1][1]) return 0;
21     return 1;
22 }
23 void flip(int i,int j) //翻棋 
24 {
25     g[i][j]^=1;
26     g[i-1][j]^=1;
27     g[i+1][j]^=1;
28     g[i][j-1]^=1;
29     g[i][j+1]^=1;
30 }
31 void dfs(int x,int y,int cnt,int n)
32 {
33     if(cnt==n){
34         flag=judge(g);
35         return;
36     }
37     if(flag || y>4) return;
38     flip(x,y);  
39     if(x<4) dfs(x+1,y,cnt+1,n);
40     else dfs(1,y+1,cnt+1,n);
41     flip(x,y);  
42     if(x<4) dfs(x+1,y,cnt,n);
43     else dfs(1,y+1,cnt,n);
44 } 
45 int main(void)
46 {
47     char c[10];
48     while(scanf("%s",c)!=EOF)
49     {
50         memset(g,0,sizeof(g));
51         for(int i=0;i<4;i++) g[1][i+1]=c[i]=='b'?1:0;
52         for(int i=1;i<4;i++){
53             scanf("%s",c);
54             for(int j=0;j<4;j++)
55             g[i+1][j+1]=c[j]=='b'?1:0;
56         }
57         flag=0;
58         int cnt=-1;
59         for(int i=0;i<16;i++){
60             dfs(1,1,0,i);
61             if(flag){
62                 cnt=i;break;
63             }
64         }
65         if(cnt==-1) puts("Impossible");
66         else printf("%d
",cnt);
67     }
68     return 0; 
69 }
70 /*
71 
72 bwwb
73 bbwb
74 bwwb
75 bwww
76 
77 bwbw 
78 bwww 
79 wwwb 
80 wwwb 
81 
82 bwww
83 wwww
84 wwww
85 wwww
86 
87 */