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 = ;

 int fac[] = {,,,,,,,,};
 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[][] = {,, -,, ,, ,- };

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

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

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

 void bfs(string str, int pos)
 {
     queue<int>que;

     //string str = "123456789";
     int i, m, n = Kt(str, );
     int ii, jj, ti, tj, t;

     nod[n].str = str;
     nod[n].postion9 = pos;
     step[n].last = -;
     que.push(n);
     used[n] = true;

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

             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 == )
                     return ;
                 que.push(m);
             }
             swap(nod[n].str[nod[n].postion9], nod[n].str[t]);
         }
     }
 }

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

 int main()
 {

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