POJ1077 八数码 BFS

BFS 几天的超时...

A*算法不会,哪天再看去了.

/*
    倒搜超时,
    改成顺序搜超时
    然后把记录路径改成只记录当前点的操作,把上次的位置记录下AC..不完整的人生啊
*/

#include <iostream>
#include <queue>
#include <vector>
#include <iterator>
#include <string>
using namespace std;

const int MAXX_SIZE = 362885;

int fac[] = {1,1,2,6,24,120,720,5040,40320};
bool used[MAXX_SIZE];

struct Nod
{
    string str;    //数组
    int postion9;  //可交换的位置
}nod[MAXX_SIZE];

struct Step
{
    int last;
    char c;        //l0, r1, u2, d3

}step[MAXX_SIZE];

int dir[][2] = {1,0, -1,0, 0,1, 0,-1 };

char st[] = {'d', 'u', 'r', 'l'};

int Kt(string str, int n)
{
    int i, j, cnt, sum=0;
    for (i=0; i<n; i++)
    {
        cnt = 0;
        for (j=i+1; j<n; j++)
            if (str[i] > str[j])
                cnt++;
        sum += cnt * fac[n-1-i];
    }
    return sum;
}

string InverKt(int sum, int n)
{
    int i, j, t;
    bool Int[20];
    string str;
    for (i=0; i<n; i++)
    {
        t = sum / fac[n-1-i];
        for (j=0; j<n; j++)
        {
            if (Int[j])
            {
                if (t == 0)
                    break;
                t--;
            }
        }
        str += j+1+'0';
        Int[j] = false;
        sum %= fac[n-1-i];
    }
    //str += '';
    return str;
}


void bfs(string str, int pos)
{
    queue<int>que;
    
    //string str = "123456789";
    int i, m, n = Kt(str, 9);
    int ii, jj, ti, tj, t;
    
    nod[n].str = str;
    nod[n].postion9 = pos;
    step[n].last = -1;
    que.push(n);
    used[n] = true;

    while (!que.empty())
    {
        n = que.front();
        que.pop();
        ii = nod[n].postion9 / 3;
        jj = nod[n].postion9 % 3;
        for (i=0; i<4; i++)
        {
            ti = ii + dir[i][0];
            tj = jj + dir[i][1];
            if (ti < 0 || ti > 2 || tj < 0 || tj > 2)
                continue;
            t = ti*3+tj;
            swap(nod[n].str[nod[n].postion9], nod[n].str[t]);
            m = Kt(nod[n].str, 9);

            if (!used[m])
            {
                used[m] = true;
                nod[m].str = nod[n].str;
                nod[m].postion9 = t;
                step[m].last = n;
                step[m].c = st[i] ;
                //step[m].str = step[n] + st[i]; 超时!
                if (m == 0)
                    return ;
                que.push(m);
            }
            swap(nod[n].str[nod[n].postion9], nod[n].str[t]);
        }
    }
}

void show(int m)
{
    if (step[m].last == -1)
        return;
    show(step[m].last);
    cout<<step[m].c;
}

int main()
{
    
    int i, n=9, m, pos;
    char c;
    string str;
    for (i=0; i<n; i++)
    {
        cin>>c;
        if (c == 'x')
        {
            pos = i;
            c = '9';
        }
        str+= c;
    }
    bfs(str, pos);
//    m = Kt(str, n);
    if (!used[0])
        cout<<"unsolvable"<<endl;
    else
        show(0);
    return 0;
}