POJ1753 filp game

（转）

感谢大神的代码，加了自己读代码时加的注释，可能更好看一点。

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:

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.

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

#include <stempempdio.h>const int inf=9999999;
char s[10];
int map[10][10],i,j;
int ans=inf;int panduan()
{int x=map[0][0];for (i=0; i<4; i++){for (j=0; j<4; j++){if (map[i][j]!=x)return 0;}}return 1;
}void fan (int x,int y)
{map[x][y]=!map[x][y];if (x - 1 >= 0)map[x-1][y]=!map[x-1][y];if (x + 1 < 4)map[x+1][y]=!map[x+1][y];if (y - 1 >= 0)map[x][y-1]=!map[x][y-1];if (y + 1 < 4)map[x][y+1]=!map[x][y+1];
}int dfs (int x,int y,int t)
{if (panduan())//先判断是不是完活状态，完活就退出本状态{if (ans > t)ans = t ;return 0;}if (x >= 4 || y >= 4)return 0;else
    {
        int nx,ny;      nx = (x + 1)%4;//应该是控制下一步的位移
        ny = y + ( x + 1 ) / 4;//同上      dfs (nx,ny,t);
        fan (x,y);//把自己翻了       dfs (nx,ny,t+1);
        fan (x,y);//把自己翻回来
    }return 0;
}int main ()
{for (i=0; i<4; i++){scanf ("%s",s);for (j=0; j<4; j++){if (s[j]=='b')map[i][j]=0;elsemap[i][j]=1;}}dfs (0,0,0);if (ans == inf )printf ("Impossible\n");else printf ("%d\n",ans);return 0;
}

自己打了一遍发现这个做法和普通的模板的题（油田）的不同在于这个需要的顺序非常关键，像油田那题需要定义一个走步的数组dx[],dy[]，而这个则是不断往右走（模4）再往下走（y+1），这样控制结束就非常简（bao）单（li）。

自己打的：

/*
ID: chun SU
TASK: test
LANG: C++
*/
#include <fstream>
#include<stdio.h>
#include<iostream>
//#include<bits/stdc++.h>
#include<string>
//#define local
using namespace std;
int temp[4][4];
bool vis[4][4]={false};
string s;
#define inf 0x3f3f3f
int ans=inf;
int  judge()
{int x = temp[0][0];for(int i=0;i<4;i++)for(int j=0;j<4;j++){if(temp[i][j]!=x)return 0;}return 1;
}
int dx[]={-1,1,0,0};
int dy[]={0,0,1,-1};
void fan(int x,int y)
{temp[x][y]=!temp[x][y];if (x - 1 >= 0)temp[x-1][y]=!temp[x-1][y];if (x + 1 < 4)temp[x+1][y]=!temp[x+1][y];if (y - 1 >= 0)temp[x][y-1]=!temp[x][y-1];if (y + 1 < 4)temp[x][y+1]=!temp[x][y+1];
}
void dfs(int x,int y,int t)
{if(judge()){ans=min(ans,t);return ;}if(y>=4) return ;else{
//      for(int k=0;k<4;k++)
//      {int nx = (x + 1)%4;//应该是控制下一步的位移int ny = y + ( x + 1 ) / 4;//同上// int  nx=x+dx[k];//    int ny=y+dy[k];
//          if(nx>=0&&ny>=0&&nx<4&&ny<4/*&&!vis[nx][ny]*/)
//          {dfs(nx,ny,t);fan(x,y);//   vis[x][y]=true;dfs(nx,ny,t+1);fan(x,y);// vis[x][y]=false;
//          }
//      }}return ;
}
int main()
{
#ifdef local
freopen("C:\\Users\\苏淳sv\\Desktop\\输入.txt","r",stdin);
#else
ofstream fout ("gift1.out");
ifstream fin ("gift1.in");
#endif
for(int i=0;i<4;i++)
{//scanf("%s",s);cin>>s;for(int j=0;j<s.size();j++){if(s[j]=='b') temp[i][j]=1;else temp[i][j]=0;}s.clear();
}
dfs(0,0,0);
if (ans == inf )printf ("Impossible\n");else printf ("%d\n",ans);return 0;
}

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