POJ 2965:The Pilots Brothers' refrigerator

这个题目的解法与poj:1753差不多，只是我添加了一个链表来记录每次操作的步骤

我就直接贴代码了

#include <iostream>
#include <queue>
#include <list>
using namespace std;

#define SIZE 4
#define	MAX	65536
int flip[]={63624,62532,61986,61713,36744,20292,12066,7953,35064,17652,8946,4593,34959,17487,8751,4383};
queue<int> myqueue;
struct
{
	int i;
	int prior;
}record[MAX];

int main()
{
	int answer[SIZE*SIZE];
	int step[MAX];
	bool flag[MAX];
	memset(step,0,MAX);
	memset(flag,false,MAX);
	int stat = 0;
	int nextstat=0;
	int i,j,tmp1;
	char c;
	for (i=0;i<SIZE*SIZE;++i)
	{
		cin>>c;
		c=='+'?(stat<<=1)++:stat<<=1;
	}
	flag[stat]=0;
	step[stat]=0;
	record[stat].i=-1;
	myqueue.push(stat);
	while(!myqueue.empty())
	{
		stat=myqueue.front();
		myqueue.pop();
		for(i=0;i<SIZE*SIZE;++i)
		{
			nextstat=stat^flip[i];
			if(flag[nextstat]==true)
				continue;
			flag[nextstat]=true;
			step[nextstat]=step[stat]+1;
			record[nextstat].prior=stat;
			record[nextstat].i=i;
			if (nextstat==0x0000)
			{
				cout<<step[nextstat]<<endl;
				tmp1=nextstat;
				answer[0]=0;
				for (j=0;j<step[nextstat]-1;j++)
				{
					tmp1=record[tmp1].prior;
					answer[j+1]=tmp1;
				}
				for (j=step[nextstat]-1;j>=0;j--)
				{
					cout<<((record[answer[j]].i)/4)+1 <<"  "<<((record[answer[j]].i)%4)+1<<endl;
				}
				return 0;
			}
			myqueue.push(nextstat);
		}
	}
	return 0;
}