POJ1077 八数码 BFS

BFS 几天的超时...

A*算法不会,哪天再看去了.

  1 /*

  2     倒搜超时,

  3     改成顺序搜超时

  4     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

  5 */

  6 

  7 #include <iostream>

  8 #include <queue>

  9 #include <vector>

 10 #include <iterator>

 11 #include <string>

 12 using namespace std;

 13 

 14 const int MAXX_SIZE = 362885;

 15 

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

 17 bool used[MAXX_SIZE];

 18 

 19 struct Nod

 20 {

 21     string str;    //数组

 22     int postion9;  //可交换的位置

 23 }nod[MAXX_SIZE];

 24 

 25 struct Step

 26 {

 27     int last;

 28     char c;        //l0, r1, u2, d3

 29 

 30 }step[MAXX_SIZE];

 31 

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

 33 

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

 35 

 36 int Kt(string str, int n)

 37 {

 38     int i, j, cnt, sum=0;

 39     for (i=0; i<n; i++)

 40     {

 41         cnt = 0;

 42         for (j=i+1; j<n; j++)

 43             if (str[i] > str[j])

 44                 cnt++;

 45         sum += cnt * fac[n-1-i];

 46     }

 47     return sum;

 48 }

 49 

 50 string InverKt(int sum, int n)

 51 {

 52     int i, j, t;

 53     bool Int[20];

 54     string str;

 55     for (i=0; i<n; i++)

 56     {

 57         t = sum / fac[n-1-i];

 58         for (j=0; j<n; j++)

 59         {

 60             if (Int[j])

 61             {

 62                 if (t == 0)

 63                     break;

 64                 t--;

 65             }

 66         }

 67         str += j+1+'0';

 68         Int[j] = false;

 69         sum %= fac[n-1-i];

 70     }

 71     //str += '\0';

 72     return str;

 73 }

 74 

 75 

 76 void bfs(string str, int pos)

 77 {

 78     queue<int>que;

 79     

 80     //string str = "123456789";

 81     int i, m, n = Kt(str, 9);

 82     int ii, jj, ti, tj, t;

 83     

 84     nod[n].str = str;

 85     nod[n].postion9 = pos;

 86     step[n].last = -1;

 87     que.push(n);

 88     used[n] = true;

 89 

 90     while (!que.empty())

 91     {

 92         n = que.front();

 93         que.pop();

 94         ii = nod[n].postion9 / 3;

 95         jj = nod[n].postion9 % 3;

 96         for (i=0; i<4; i++)

 97         {

 98             ti = ii + dir[i][0];

 99             tj = jj + dir[i][1];

100             if (ti < 0 || ti > 2 || tj < 0 || tj > 2)

101                 continue;

102             t = ti*3+tj;

103             swap(nod[n].str[nod[n].postion9], nod[n].str[t]);

104             m = Kt(nod[n].str, 9);

105 

106             if (!used[m])

107             {

108                 used[m] = true;

109                 nod[m].str = nod[n].str;

110                 nod[m].postion9 = t;

111                 step[m].last = n;

112                 step[m].c = st[i] ;

113                 //step[m].str = step[n] + st[i]; 超时!

114                 if (m == 0)

115                     return ;

116                 que.push(m);

117             }

118             swap(nod[n].str[nod[n].postion9], nod[n].str[t]);

119         }

120     }

121 }

122 

123 void show(int m)

124 {

125     if (step[m].last == -1)

126         return;

127     show(step[m].last);

128     cout<<step[m].c;

129 }

130 

131 int main()

132 {

133     

134     int i, n=9, m, pos;

135     char c;

136     string str;

137     for (i=0; i<n; i++)

138     {

139         cin>>c;

140         if (c == 'x')

141         {

142             pos = i;

143             c = '9';

144         }

145         str+= c;

146     }

147     bfs(str, pos);

148 //    m = Kt(str, n);

149     if (!used[0])

150         cout<<"unsolvable"<<endl;

151     else

152         show(0);

153     return 0;

154 }

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