【hdu 1067】Gap

Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 1123 Accepted Submission(s): 595

Problem 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 open space at the left end of the rows: “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

【题目链接】:http://acm.hdu.edu.cn/showproblem.php?pid=1067

【题解】



搜索题。

这题的判重方法和八数码类似;

把整张4*8的图降成一维的图;

判重的时候转换成一个字符串判重就好;

在bfs的队列里面

记录

{

———-4*7个数码在哪一个位置;

———-4个空格的位置;

———-当前的步数;

———-这张图用一维字符串的表示;

}

在扩展的时候

枚举4个空格.

看看4个空格左边是什么;

如果是0或者尾数为7就跳过;

否则找到比它大1的数码(我们有记录)的位置;

(二维的坐标和一维的坐标可以通过(x-1)*8+y来转换)

然后调换字符串中两个数目;

更改那个被调换的数目的位置(不要忘了！)

更改空格的位置;

修改当前步数;

入队。

如此循环一下就OK了.

判重用map



【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define rei(x) scanf("%d",&x)
#define rel(x) scanf("%I64d",&x)
typedef pair<int,int> pii;
typedef pair<LL,LL> pll;
const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const int goal[32] = {11,12,13,14,15,16,17,0,21,22,23,24,25,26,27,0,31,32,33,34,35,36,37,0,41,42,43,44,45,46,47,0};
const double pi = acos(-1.0);
const int MAXN = 110;
struct abc
{
    pii po[50],kong[5];
    int dis;
    string s;
};
char chushi[5][10];
string sgoal,ts;
map <string,int> dic;
queue <abc> dl;
bool bfs()
{
    dic.clear();
    while (!dl.empty()) dl.pop();
    abc pre;
    pre.dis = 0;pre.s = " ";
    int num = 0;
    rep1(i,1,4)
    {
        rep1(j,1,8)
            {
                pre.s+=chushi[i][j];
                if (chushi[i][j]==0)
                    pre.kong[++num] = {i,j};
                else
                    pre.po[int(chushi[i][j])] = {i,j};
            }
    }
    dic[pre.s] = 1;
    if (dic[sgoal])
    {
        puts("0");
        return true;
    }
    dl.push(pre);
    while (!dl.empty())
    {
        abc temp = dl.front();dl.pop();
        for (int i = 1;i <= 4;i++)
        {
            abc t = temp;
            int tx = t.kong[i].fi,ty = t.kong[i].se;
            int sz = (tx-1)*8+ty;
            int judge = t.s[sz-1];
            if (judge==0 || judge%10==7) continue;
            int tx1 = t.po[judge+1].fi,ty1 = t.po[judge+1].se;
            int sz1 = (tx1-1)*8+ty1;
            swap(t.s[sz],t.s[sz1]);
            if (!dic[t.s])
            {
                dic[t.s] = 1;
                t.kong[i] = {tx1,ty1};
                t.po[judge+1] = {tx,ty};
                t.dis++;
                if (dic[sgoal])
                {
                    printf("%d\n",t.dis);
                    return true;
                }
                dl.push(t);
            }
        }
    }
    return false;
}
int main()
{
    //freopen("F:\\rush.txt","r",stdin);
    sgoal = " ";
    for (int i = 0;i <= 31;i++)
        sgoal+=goal[i];
    int T;
    rei(T);
    while (T--)
    {
        memset(chushi,0,sizeof chushi);
        rep1(i,1,4)
            rep1(j,2,8)
                {
                    int x;
                    rei(x);
                    chushi[i][j] = x;
                    if (x%10==1)
                        swap(chushi[x/10][1],chushi[i][j]);
                }
        if (!bfs())
            puts("-1");
    }
    return 0;
}