poj 1753 Flip Game

Flip Game

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 30395 Accepted: 13199

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

题目大意：

给你一个4X4的方格，里面是w或者是b，现在你翻转其中某一个棋子的话棋子以及棋子的四周也会翻转，问能否达到全部的黑或者白。经典的DFS题目，复习一下........

题目不是很难，在dfs的时候要考虑好访问的顺序，按照这个顺序来判断跳出的条件。什么时候回溯也要有判断。

#include<iostream>#include<stdio.h>using namespace  std;char mp[6][6];bool flag;int step;void in(){    for(int i=1;i<=4;i++)    {        for(int j=1;j<=4;j++)        scanf("%c",&mp[i][j]);        getchar();    }}bool judge(){    bool flag1=true,flag2=true;    for(int i=1;i<=4;i++)        for(int j=1;j<=4;j++)        if(mp[i][j]!='b') flag1=false;    for(int i=1;i<=4;i++)        for(int j=1;j<=4;j++)        if(mp[i][j]!='w') flag2=false;        return flag1||flag2;}int xx[5]={0,0,0,-1,1};int yy[5]={0,1,-1,0,0};void flip(int row,int col){    for(int i=0;i<=4;i++)     if(mp[row+xx[i]][col+yy[i]]=='b') mp[row+xx[i]][col+yy[i]]='w'; else mp[row+xx[i]][col+yy[i]]='b';}void dfs(int row,int col,int deep){    //printf("%d %d %d\n",row,col,deep);    if(deep==step) {flag=judge();return;}    if(flag||row==5) return;    flip(row,col);    if(col<4) dfs(row,col+1,deep+1); //遍历的过程是向下向右    else      dfs(row+1,col,deep+1);    flip(row,col);                   //若是没有成功，则回溯    if(col<4) dfs(row,col+1,deep);   //左向右遍历，若是col==4 rol变成1，row++也就是说重新一行开始    else      dfs(row+1,1,deep);    return;}int main (){    in();    for(step=0;step<=16;step++)  //对每一步产生的可能性进行枚举    {                            //至于为什么是16，考虑到4x4=16格，而每一格只有黑白两种情况，则全部的可能性为2^16        dfs(1,1,0);        if(flag) break;    }    if(flag) cout<<step<<endl;    else     cout<<"Impossible"<<endl;}