POJ 1753 Flip Game（dfs+枚举）

Flip Game

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 34269		Accepted: 14969

DescriptionFlip 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: Choose any one of the 16 pieces. 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. InputThe input consists of 4 lines with 4 characters "w" or "b" each that denote game field position.OutputWrite 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

4*4的棋盘，棋子有黑白两色，翻一个棋子，其四周也翻过来，问最少多少次全部变成黑色或全部白色，如若不能输出"Impossible"

直接枚举0~16中的所有情况 dfs看是否满足

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <stdlib.h>#define N 5#define inf 0x3f3f3f3fusing namespace std;bool a;bool flag;int xx[] = {0,0,-1,1};int yy[] = {1,-1,0,0};int ans;bool judge()               //检查棋子是否全白 或 全黑{    for(int i=1;i<5;i++)    {        for(int j=1;j<5;j++)          {              if(a[1][1]!=a[i][j])              return false;          }    }    return true;}void change(int x,int y)         //翻四周和自身{    for(int i=0;i<4;i++)    {        int c=x+xx[i];        int d=y+yy[i];        if(c>0 && c<5 && d>0 && d<5)        {            if(a[c][d]) a[c][d]=false;            else a[c][d] = true;        }    }     if(a[x][y]) a[x][y]=false;            else a[x][y] = true;}void dfs(int x,int y,int step){    if(step == ans)               {        if(judge())         //满足标记直接退出          flag=true;        return ;    }    if(x==5 || flag) return ;      change(x,y);                        if(y<4)                  //遍历所有的点                      dfs(x,y+1,step+1);    else dfs(x+1,1,step+1);     change(x,y);                 if(y<4)     dfs(x,y+1,step);     else dfs(x+1,1,step);}int main(){    memset(a,false,sizeof(a));    for(int i=1;i<=4;i++)      {          for(int j=1;j<=4;j++)             {                 char s;                 cin>>s;                 if(s=='b') a[i][j]=true;             }      }      flag=false;      for(ans=0;ans<=16;ans++)       // 枚举所有的可能      {           dfs(1,1,0);           if(flag)             break;      }      if(flag)      printf("%d\n",ans);      else printf("Impossible\n");    return 0;}