[sicily online]1048. Inverso

Constraints

Time Limit: 1 secs, Memory Limit: 32 MB

Description

The game of ‘Inverso’ is played on a 3x3 grid of colored fields (each field is either black or white). Each field is numbered as follows: 
1  2  3 
4  5  6 
7  8  9 
The player can click on a field, which will result in the inversion of that field and all its direct neighboring fields (i.e. the color changes from black to white or vice-versa). Hence, 
Clicking field 1 inverts fields 1, 2, 4, 5 
Clicking field 2 inverts fields 1, 2, 3, 4, 5, 6 
Clicking field 3 inverts fields 2, 3, 5, 6 
Clicking 4 inverts fields 1, 2, 4, 5, 7, 8 
Clicking 5 inverts fields all fields 
Clicking 6 inverts fields 2, 3, 5, 6, 8, 9 
Clicking 7 inverts fields 4, 5, 7, 8 
Clicking 8 inverts fields 4, 5, 6, 7, 8, 9 
Clicking 9 inverts fields 5, 6, 8, 9 
The aim of the game is to find the shortest sequence of clicks to make all fields white from a given start coloring of the grid. 

Input

The first line contains a number N (0≤N≤10000) of runs. The following N lines each contain a string of nine letters ‘b’ (black) or ‘w’ (white) for the color of the fields 1 to 9. This is the initial coloring of the grid.

Output

For each input string the shortest word in the letters ‘1’… ‘9’ (for clicking field one, …, nine) which makes all fields white. If there is more than one shortest word then the lexicographically smallest one must be printed (‘1234’ is smaller than ‘1342’).

Sample Input

3
bbwbwbwbw
bwwwbwbwb
bbbbwbbbw

Sample Output

2459 
267
356789

题目分析：

像图那样进行广度遍历，但是要设个标记，遍历过的就不要再遍历了，还要注意wwwwwwwww输出11，这个调了半个小时 郁闷

/*
*/
#include<iostream>
#include <iomanip>
#include<stdio.h>
#include<cmath>
#include<iomanip>
#include<list>
#include <map>
#include <vector>
#include <string>
#include <algorithm>
#include <sstream>
#include <stack>
#include<queue>
#include<string.h>
using namespace std;

int data[9][10]={{1,2,4,5,0},{1,2,3,4,5,6,0},{2,3,5,6,0},{1,2,4,5,7,8,0},{1,2,3,4,5,6,7,8,9,0},{2,3,5,6,8,9,0},{4,5,7,8,0},{4,5,6,7,8,9,0},{5,6,8,9,0}};

typedef struct STATUS
{
	unsigned int a;
	vector<int> record;
}status;

inline bool judge(int a)
{
	if(a==0)
		return true;
	return false;
}

int main()
{
	int n;
	cin>>n;
	for(int xx=0;xx<n;xx++)
	{
		unsigned int gra=0;
		for(int i=0;i<9;i++)//w是0，b是1
		{
			char x;
			cin>>x;
			if(x=='b')
				gra+=pow(2.0,i);
		}
		if(gra==0)
		{
			cout<<11<<endl;
			continue;
		}
		status tmp;
		tmp.a=gra;
		bool flag[513];
		memset(flag,0,sizeof(flag));
		queue<status> qe;
		qe.push(tmp);
		flag[gra]=true;
		while(1)
		{
			status wh=qe.front();
			flag[wh.a]=true;
			qe.pop();
			if(judge(wh.a))
			{
				for(vector<int>::size_type i=0;i<wh.record.size();i++)
					cout<<wh.record[i]+1;
				cout<<endl;
				break;
			}
			for(int i=0;i<9;i++)
			{
				status tt;
				tt.a=wh.a;
				tt.record=wh.record;

				for(int j=0;j<9&&data[i][j]!=0;j++)
				{
					unsigned int qw=pow(2.0,data[i][j]-1);
					tt.a=tt.a^qw;
				}//end for j
				if(flag[tt.a]==true)
					continue;
				tt.record.push_back(i);
				qe.push(tt);
			}//end for i
		}//end while
	}
}