2046 Gap poj

Gap

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 1595		Accepted: 724

Description

Let's play a card game called Gap. 
You have 28 cards labeled with two-digit numbers. The first digit (from 1 to 4) represents the suit of the card, and the second digit (from 1 to 7) represents the value of the card.

First, you shu2e the cards and lay them face up on the table in four rows of seven cards, leaving a space of one card at the extreme left of each row. The following shows an example of initial layout. 

Next, you remove all cards of value 1, and put them in the Z喎�"https://www.2cto.com/kf/ware/vc/" target="_blank" class="keylink">vcGVuIHNwYWNlIGF0IHRoZSBsZWZ0IGVuZCBvZiB0aGUgcm93czog"11" to the top row, "21" to the next, and so on. 

Now you have 28 cards and four spaces, called gaps, in four rows and eight columns. You start moving cards from this layout. 

At each move, you choose one of the four gaps and fill it with the successor of the left neighbor of the gap. The successor of a card is the next card in the same suit, when it exists. For instance the successor of "42" is "43", and "27" has no successor. 

In the above layout, you can move "43" to the gap at the right of "42", or "36" to the gap at the right of "35". If you move "43", a new gap is generated to the right of "16". You cannot move any card to the right of a card of value 7, nor to the right of a gap. 

The goal of the game is, by choosing clever moves, to make four ascending sequences of the same suit, as follows. 

Your task is to find the minimum number of moves to reach the goal layout.

Input

The input starts with a line containing the number of initial layouts that follow. 

Each layout consists of five lines - a blank line and four lines which represent initial layouts of four rows. Each row has seven two-digit numbers which correspond to the cards. 

Output

For each initial layout, produce a line with the minimum number of moves to reach the goal layout. Note that this number should not include the initial four moves of the cards of value 1. If there is no move sequence from the initial layout to the goal layout, produce "-1".

Sample Input

4

12 13 14 15 16 17 21
22 23 24 25 26 27 31
32 33 34 35 36 37 41
42 43 44 45 46 47 11

26 31 13 44 21 24 42
17 45 23 25 41 36 11
46 34 14 12 37 32 47
16 43 27 35 22 33 15

17 12 16 13 15 14 11
27 22 26 23 25 24 21
37 32 36 33 35 34 31
47 42 46 43 45 44 41

27 14 22 35 32 46 33
13 17 36 24 44 21 15
43 16 45 47 23 11 26
25 37 41 34 42 12 31

Sample Output

0
33
60
-1

/*
BFS,利用set来判重.将状态封装到string大大简化了处理过程
*/
#include <iostream>
#include <string>
#include <queue>
#include <set>

using namespace std;

set<string> hashSets;

struct elem
{
    int steps;
    string state;
};
queue<elem> bfsq;

//最终状态++表示空格
string finalState = "11121314151617++21222324252627++31323334353637++41424344454647++";
//记录初始状态
string firstState = "";
int minSteps;

//在当前状态中寻找二位数num
int findNumPos(const string &state, const string &num)
{
    for(int i = 0; i < 32; i++)
    {
        if(state[i * 2] == num[0] && state[i * 2 + 1] == num[1])
            return i * 2;
    }
    return -1;
}

//广搜,需要判重
void bfs()
{
    hashSets.clear();
    hashSets.insert(firstState);
    while(!bfsq.empty()) bfsq.pop();
    elem curelem, newelem;
    curelem.steps = 0;
    curelem.state = firstState;
    bfsq.push(curelem);

    string curstate, newstate, num1, num2;
    int p1, p2, cursteps, i;
    while(!bfsq.empty())
    {
        curelem = bfsq.front();
        bfsq.pop();
        cursteps = curelem.steps;
        curstate = curelem.state;

        //得到最优解
        if(curstate == finalState)
        {
            minSteps = cursteps;
            return;
        }
        int startPos = 0;

        //从四个空格处分别取
        for(i = 0; i < 4; i++)
        {
            p1 = curstate.find_first_of('+', startPos);
            
            num1 = curstate[p1 - 2];
            num1 += curstate[p1 - 1];
            
            //当前空格可取的条件是有数字邻居且数字邻居的第二位不是7
            if(num1[0] != '+' && num1[1] != '7') 
            {
                num2 = num1;
                num2[1]++;

                //找到后继的位置
                p2 = findNumPos(curstate, num2);
                newstate = curstate;

                //交换后继与空格
                newstate[p1] = curstate[p2];
                newstate[p1 + 1] = curstate[p2 + 1];
                newstate[p2] = newstate[p2 + 1] = '+';
                
                //判断当前状态是否需要进队列
                if(hashSets.find(newstate) == hashSets.end() && cursteps + 1 < minSteps)
                {
                    hashSets.insert(newstate);
                    newelem.steps = cursteps + 1;
                    newelem.state = newstate;
                    bfsq.push(newelem);
                }
                
                
            }
            startPos = p1 + 2;
        }
    }
}
int main()
{
    int caseN, i, j;
    string temp;
    cin>>caseN;
    while(caseN--)
    {
        //初始化
        minSteps = INT_MAX;
        firstState = finalState;
        //将初始状态的四行的初始列置为空格
        firstState[0] = firstState[1] = firstState[16] = firstState[17] = firstState[32] = firstState[33] = firstState[48] = firstState[49] = '+';
        for(i = 0; i < 4; i++)
        {
            for(j = 1; j <= 7; j++)
            {
                cin>>temp;
                //如果是尾1二位数，需要放到对应行的首部
                if(temp[1] == '1')
                {
                    int pos = (int(temp[0] - '0') - 1) * 16;
                    firstState[pos] = temp[0];
                    firstState[pos + 1] = temp[1];
                    firstState[i * 16 + j * 2] = firstState[i * 16 + j * 2 + 1] = '+';
                }
                else
                {
                    firstState[i * 16 + j * 2] = temp[0];
                    firstState[i * 16 + j * 2 + 1] = temp[1];
                }
            }
        }
        bfs();
        if(minSteps == INT_MAX) cout<<"-1"<<endl;
        else cout<<minSteps<<endl;
    }
    return 0;
}