flip game

Flip Game

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 27237		Accepted: 11816

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

思路：很简单就是广搜，每个点过后找下一种情况进入队列，技巧是标记

位运算：参考资料：http://www.cnblogs.com/911/archive/2008/05/20/1203477.html

“异或”运算符（^）
他的规则是：若参加运算的两个二进制位值相同则为0，否则为1
即0∧0=0，0∧1=1，1∧0=1， 1∧1=0
    例：   00111001
        ∧ 00101010
           00010011 
c语言源代码：
#include <stdio.h>
main()
{
 int a=071;
 int b = 052;
 printf("%d",a^b);
}
应用：
使特定位翻转，与1异或
设有数01111010（2），想使其低4位翻转，即1变0，0变1.可以将其与00001111（2）进行“异或”运算，

即：
 01111010
^00001111
 01110101
运算结果的低４位正好是原数低４位的翻转。可见，要使哪几位翻转就将与其进行∧运算的该几位置为１

即可。

             

#include<stdio.h>
#include<iostream>
#include<queue>
#define N 4
using namespace std;;
typedef struct{
	int step;
	int state;
}node; 
queue<node> q;
int d[4][2] = {{0,1}, {1,0}, {-1,0}, {0,-1}};
int a[16] = {0};
int start;
int visited[70000];//2的16次方65536

void inita(){
	int i, j, k ;
	int x, y;
	int count = 0; 
	int temp;
	for(i = 0; i < N; i++)
		for(j = 0; j < N; j++){//给每个状态赋值，一共是2^16个状态
			temp = 0;
			temp = 1 << (3-i) * 4 + (3-j);//给后面的行，列填零
			for(k = 0; k < 4; k++){
				x = i + d[k][0];
				y = j + d[k][1]; 	
				if(x >= 0 && y >= 0 && x < N && y < N)
					temp = temp ^ (1 << (3-x)*4 + (3-y));//某位数及其会翻转的结果
			}
			a[count++] = temp;
		}
}

void read_data(){
	int i, j;
	char str[N];
	start = 0;
	for(i = 0; i < N; i++){
		scanf("%s", str);
		for(j = 0; j < N; j++){
			start <<= 1;
			if(str[j] == 'b')
				start += 1;
		}
	}
}

void bfs(){
	node curr, next;
	int i;

	while(!q.empty()){
		curr = q.front();
		q.pop();
	//	printf("%d %x\n", curr.step , curr.state);
	//	getchar();
		if(curr.state == 0xffff || curr.state == 0){
			printf("%d\n", curr.step);
			return;
		}
		for(i = 0; i < N * N; i++){
			next.state = curr.state ^ a[i];
			if(visited[next.state] == 0){
				visited[next.state] = 1;
				next.step = curr.step + 1;
				q.push(next);
			}
		}
	}
	printf("Impossible\n");
}

int main(){
	node b;
	inita();
	read_data();
	while(!q.empty())
		q.pop();
	b.step = 0;
	b.state = start;
	q.push(b);
	bfs();
	return 0;
}