POJ-1753(DFS,Notation,Bit operation,Queue)
1753:Flip Game
时间限制:
1000ms
内存限制:
65536kB
描述
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.
输入
The input consists of 4 lines with 4 characters "w" or "b" each that denote game field position.
输出
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).
样例输入
bwwb
bbwb
bwwb
bwww
样例输出
4
#include"iostream"
#include"queue"
using namespace std;
const int MAX=0xffff;//即2^16,分别对应状态id从0到1111111111111111
int for_reverse[4][4]=
{0xc800,0xe400,0x7200,0x3100,
0x8c80,0x4e40,0x2720,0x1310,
0x08c8,0x04e4,0x0272,0x0131,
0x008c,0x004e,0x0027,0x0013};
int main()
{
char input;
int id=0;
for(int i=0;i<4;i++)
for(int j=0;j<4;j++)
{
cin>>input;
if(input=='w')
{
id<<=1;
}
else if(input=='b')
{
id<<=1;
id++;
}
}
bool done[MAX]={false};//记录状态id=i是否遍历过
int step[MAX];//记录到达状态id=i需要的次数
done[id]=true;//初始点标记为遍历过
step[id]=0;//初始状态对应id次数
queue<int> st;
st.push(id);
if(id==0||id==0xffff)
{cout<<0<<endl;return 0;} //如果初始状态已经满足要求直接输出
while(!st.empty())//利用栈DFS遍历不同id值
{
int temp=st.front();
st.pop();
id=temp;
//cout<<"id:"<<id<<endl;
//cout<<"step[id]"<<step[id]<<endl;
for(int i=0;i<4;i++)//遍历16种操作
for(int j=0;j<4;j++)
{
temp=id;//注意了,每次都要重新赋值
temp^=for_reverse[i][j];//利用异或符翻转
if(temp==0||temp==0xffff)
{
cout<<step[id]+1<<endl;
return 0;
}
else if(done[temp]==false)
{
step[temp]=step[id]+1;
done[temp]=true;
st.push(temp);
}
}//结束操作遍历
}//结束状态值id遍历
cout<<"Impossible"<<endl;
return 0;
}
