POJ

Flip Game

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 45588 Accepted: 19531

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

bwwbbbwbbwwbbwww

Sample Output

4

题意：

        一个棋盘上有黑白两种棋子，flip一颗棋子，它本身和它四周的棋子都变为相反的颜色。求最少要几步才能使棋盘上的棋子颜色全部一样。

题解：

        用dfs。逐点搜索，每搜到一个点就有两种状态，flip这个棋子或者不flip这个棋子。如果flip这个棋子，则step+1，再搜下一个点；否则step不变，直接搜下一个点。而计算下一个点的坐标可以用公式：ty=(y*4+x+1)/4;  tx=(y*4+x+1)%4; 也就是先将坐标一维展开，+1，再转化成二维。

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;char str[5][5];int map[4][4],minn;void reverse(int y,int x){    if(y-1>=0)        map[y-1][x]=!map[y-1][x];    if(x-1>=0)        map[y][x-1]=!map[y][x-1];    if(y+1<4)        map[y+1][x]=!map[y+1][x];    if(x+1<4)        map[y][x+1]=!map[y][x+1];    map[y][x]=!map[y][x];}int check(){    for(int i=0; i<4; i++)        for(int j=0; j<4; j++)            if(map[i][j]!=map[0][0])                return 0;    return 1;}void dfs(int y,int x,int step){    int ty=(y*4+x+1)/4;    int tx=(y*4+x+1)%4;    if(y==4)    {        if(check())            minn=min(minn,step);        return ;    }    reverse(y,x);    dfs(ty,tx,step+1);    reverse(y,x);    dfs(ty,tx,step);}int main(){    while(~scanf("%s",str[0]))    {        for(int i=1; i<4; i++)            scanf("%s",str[i]);        for(int i=0; i<4; i++)        {            for(int j=0; j<4; j++)            {                if(str[i][j]=='b')                    map[i][j]=1;                else                    map[i][j]=0;            }        }        minn=9999;        dfs(0,0,0);        if(minn<=16)            printf("%d\n",minn);        else            printf("Impossible\n");    }    return 0;}