Ural 1060. Flip Game dfs

1060. Flip Game

Time limit: 2.0 second
Memory limit: 64 MB

Flip game is played on a rectangular 4×4 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:

bwbwwwwwbbwbbwwb

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:

bwbwbwwwwwwbwwwb

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 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

inputoutput

bwbwwwwwbbwbbwwb

Impossible

Problem Source: 2000-2001 ACM Northeastern European Regional Programming Contest 

/**< 代码我不能再精简了 */#include <iostream>using namespace std;int a[6][6];string str[5];int minn=0xfffffff;bool check(){    int i,j,n;    n=a[1][1];    for(i=1; i<=4; i++)        for(j=1; j<=4; j++)            if(a[i][j]!=n)                return false ;    return true;}void flip(int x,int y){    a[x][y]=1-a[x][y];    a[x-1][y]=1-a[x-1][y];    a[x+1][y]=1-a[x+1][y];    a[x][y-1]=1-a[x][y-1];    a[x][y+1]=1-a[x][y+1];    return ;}void dfs(int pos,int num){    if(pos>17||pos>minn) return ;//棋子超过16个或者走步数已经超过了最小值的直接结束    if(check())    {        if(num<minn)            minn=num;        return ;    }    int x,y;    x=pos/4+1;    y=pos%4;    if(y==0)    {        x--;        y=4;    }    dfs(pos+1,num);    flip(x,y);    dfs(pos+1,num+1);    flip(x,y);    return ;}int main(){    int i,j;    for(i=1; i<=4; i++)    {        cin>>str[i];        str[i]=" "+str[i];        for(j=1; j<=4; j++)            if(str[i][j]=='b')                a[i][j]=1;            else a[i][j]=0;    }    dfs(1,0);    if(minn<17)        cout<<minn<<endl;    else        cout<<"Impossible"<<endl;}