poj 1753【枚举+dfs（位向量法）】

Flip Game

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 46519		Accepted: 19915

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：给你一个4*4的棋盘，黑棋是‘b’,白棋是‘w’，问最少翻转多少次，可以使棋盘颜色统一。棋子有黑白两面可以翻转，翻转一个棋子，周围的棋子都要翻转。
思路：《算法竞赛入门经典》上提到枚举子集的方法之一---位向量法。这种方法就是把棋盘所有可能的翻转子集情况枚举一遍，如果在翻转过程中，棋盘颜色得以统一，就将现在的步数与最小值进行比较，如果小于最小值，就更新它。

#include<stdio.h>
#define inf 0x3f3f3f3f
int map[5][5],min,n;

int Check(int map[][5])//检查棋盘颜色是否统一 
{
    int i,j;
    for(i = 0; i < 4; i ++)
        for(j = 0; j < 4; j ++)
            if(map[0][0] != map[i][j])
                return 0;
    return 1;
}

void Reverse(int num)//翻转棋子 
{
    int x,y;
    x = num/4;
    y = num%4;
    map[x][y]=!map[x][y];
    if(x>0)    map[x-1][y]=!map[x-1][y];
    if(x<3)    map[x+1][y]=!map[x+1][y];
    if(y>0)    map[x][y-1]=!map[x][y-1];
    if(y<3)    map[x][y+1]=!map[x][y+1];
    return;
}

void dfs(int num,int step)
{
    if(Check(map))
    {
        if(step < min)
            min = step;
        return;
    }
    if(num >= n)
        return;
    
    dfs(num+1,step); 
    Reverse(num);//翻转棋子 
    
    dfs(num+1,step+1);//step记录已经翻转过的棋子 
    Reverse(num);//回溯 
}

int main()
{
    int i,j;
    char ch;
    for(i = 0; i < 4; i ++)
    {
        for(j = 0; j < 4; j ++)
        {
            scanf("%c",&ch);
            if(ch == 'b')//构建棋盘 
                map[i][j] = 0;
            else
                map[i][j] = 1;
        }
        getchar();
    }
    n = 16;
    min = inf;
    dfs(0,0);
    if(min == inf)
        printf("Impossible\n");
    else
        printf("%d\n",min);
    return 0;
}