POJ1753(枚举)

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:
1 Choose any one of the 16 pieces.
2 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
4

题意

给了你一个4*4的棋盘，棋盘上的每一个格子里都有一颗一面黑一面白的棋子，一开始给出你棋盘的初始状态，你可以给任何一个棋子翻面，但是如果你翻了某一个棋子，那他上下左右的棋子都要翻面。问你最少需要翻多少次才能让16枚棋子都为黑或都为白。

思路

对一个格子操作偶数次等于没有操作，操作奇数次等于操作一次，所以答案在0~16以及impossible之间。从n=0开始枚举n次操作可能的组成情况，即操作哪几个格子，若某种组合能变图案为纯色则停止。
由于总组合数达到2^16,枚举组合情况可以用dfs来进行回溯枚举。

代码

public class FlipGame {int[][] checkerboard = new int[6][6];int count = 33;public static boolean check(int[][] checkerboard) {int result = 0;for (int i = 0; i < checkerboard.length; i++) {for (int j = 0; j < checkerboard[i].length; j++) {result = result + checkerboard[i][j];}}if (result == -16 || result == 16) {return true;} else {return false;}}public void flip(int s) {int x = s / 4 + 1;int y = s % 4 + 1;checkerboard[x][y] = -checkerboard[x][y];checkerboard[x - 1][y] = -checkerboard[x - 1][y];checkerboard[x][y - 1] = -checkerboard[x][y - 1];checkerboard[x + 1][y] = -checkerboard[x + 1][y];checkerboard[x][y + 1] = -checkerboard[x][y + 1];}public void dfs(int s,int b){//进行深搜.s代表当前的方格，b代表翻转的方格数if(check(this.checkerboard)){//如果是同一种颜色if (this.count > b){this.count = b;return;}}if(s >=16){return;}dfs(s+1,b);this.flip(s);dfs(s+1,b+1);this.flip(s);}public static void main(String[] args) {FlipGame fg = new FlipGame();Scanner sc = new Scanner(System.in);for (int i = 1; i < fg.checkerboard.length - 1; i++) {String str = sc.next();for (int j = 1; j < fg.checkerboard[i].length - 1; j++) {if (str.charAt(j - 1) == 'w') {fg.checkerboard[i][j] = 1;} else {fg.checkerboard[i][j] = -1;}}}fg.dfs(0,0);if(fg.count == 33){System.out.println("Impossible");}else {System.out.println(fg.count);}sc.close();}
}

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POJ1753(枚举)

POJ1753(枚举)

Flip Game

题意

思路

代码

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