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
Output
Sample Input
bwwb
bbwb
bwwb
bwww
Sample Output
4
题意
一个4*4黑白棋盘，每次可以将一个格子和其周围的4个格子反转成相反的颜色，问：至少反转多少次，整个棋盘一种颜色。
题解
//方法一 dfs 暴力枚举 枚举每个格子翻或不翻，如果最后同色，看下操作次数，更新答案 //方法二⭐️ 二进制枚举 & 状态压缩
Code
1 #include<cstdio> 2 #include<cstring> 3 4 int ret = 20, dir[5][2]={0,0,0,1,1,0,0,-1,-1,0}; 5 char str[5][5]; 6 7 bool check(){ 8 char c = str[0][0]; 9 for(int i = 0; i < 4; i++) 10 for(int j = 0; j < 4; j++) 11 if(str[i][j] != c) 12 return 0; 13 return 1; 14 } 15 16 void change(int x, int y){ 17 for(int i = 0; i < 5; i++){ 18 int tx = x + dir[i][0]; 19 int ty = y + dir[i][1]; 20 if(tx < 0 || tx >= 4 || ty < 0 || ty >= 4) 21 continue; 22 if(str[tx][ty] == 'w') 23 str[tx][ty] = 'b'; 24 else if(str[tx][ty] == 'b') 25 str[tx][ty] = 'w'; 26 } 27 } 28 29 void dfs(int x,int y,int step){ 30 if(step > 16) 31 return; 32 if(check()){ 33 if(step < ret) 34 ret = step; 35 return; 36 } 37 if(y >= 4){ 38 x++; 39 y=0; 40 } 41 if(x >= 4) 42 return; 43 for(int i = 0; i < 2; i++){ 44 if(i == 0){ 45 change(x, y); 46 dfs(x, y+1, step+1); 47 change(x, y); 48 } 49 else 50 dfs(x, y+1, step); 51 } 52 } 53 int main(){ 54 for(int i = 0; i < 4; i++) 55 scanf("%s", str[i]); 56 dfs(0, 0, 0); 57 if(ret <= 16) 58 printf("%d\n", ret); 59 else 60 printf("Impossible\n"); 61 return 0; 62 }
1 #include<iostream> 2 #include<cstdio> 3 using namespace std; 4 5 char mp[10][10]; 6 int m[10][10]; 7 int ret = 20; 8 9 bool check(){ 10 for(int i = 0; i < 4; i++){ 11 for(int j = 0; j < 4; j++){ 12 if(m[i][j] != m[0][0]) return false; 13 } 14 } 15 return true; 16 } 17 18 void change(int dep){ 19 int x = dep / 4; 20 int y = dep%4; 21 m[x][y] ^= 1; 22 if(x-1 >= 0) m[x-1][y] ^= 1; 23 if(x+1 < 4) m[x+1][y] ^= 1; 24 if(y-1 >= 0) m[x][y-1] ^= 1; 25 if(y+1 < 4) m[x][y+1] ^= 1; 26 } 27 28 int main(){ 29 int n; 30 scanf("%d", &n); 31 ret = 20; 32 for(int i = 0; i < 4; i++) scanf("%s", mp[i]); 33 for(int i = 0; i < 4; i++){ 34 for(int j = 0; j < 4; j++){ 35 if(mp[i][j] == 'b') m[i][j] = 0; 36 else m[i][j] = 1; 37 } 38 } 39 for(int mask = 0; mask < (1 << 16); mask++){ 40 int used = 0; 41 for(int i = 0; i < 16; i++){ 42 if(mask & (1 << i)){ 43 change(i); 44 used++; 45 } 46 } 47 if(check()){ 48 if(used < ret){ 49 ret = used; 50 } 51 } 52 for(int i = 0; i < 16; i++){ 53 if(mask & (1 << i)){ 54 change(i); 55 } 56 } 57 } 58 if(ret > 16){ 59 printf("Impossible\n"); 60 } 61 else printf("%d\n",ret); 62 return 0; 63 }