poj 1077 八数码(BFS+康托展开)

  1 /*
  2 题意：八数码问题，给出3*3的矩阵含1~8以及x，给出一个符合的解使得移动后的矩阵的顺序为1~8，最后为x
  3 
  4 题解：BFS
  5 需要用到康托展开来表示状态，不然数组无法完全表示所有状态，这样BFS就无法判断找不到解的情况(status
  6 的0ms,0KB究竟是怎么做到的，简直不能想象=。=)
  7 */
  8 #include 
  9 #include 
 10 #include 
 11 #include 
 12 #include 
 13 
 14 using namespace std;
 15 
 16 int  fac[] = {1,1,2,6,24,120,720,5040,40320,362880}; //i的阶乘为fac[i]
 17 /*  康托展开.
 18 {1...n}的全排列由小到大有序，s[]为第几个数  */
 19 int KT(int n, int s[])
 20 {
 21     int i, j, t, sum;
 22     sum = 0;
 23     for (i=0; i)
 24     {
 25         t = 0;
 26         for (j=i+1; j)
 27             if (s[j] < s[i])
 28                 t++;
 29         sum += t*fac[n-i-1];
 30     }
 31     return sum+1;
 32 }
 33 
 34 /*  康托展开的逆运算.
 35 {1...n}的全排列，中的第k个数为s[]  */
 36 void invKT(int n, int k, int s[])
 37 {
 38     int i, j, t, vst[10]={0};
 39     k--;
 40     for (i=0; i)
 41     {
 42         t = k/fac[n-i-1];
 43         for (j=1; j<=n; j++)
 44             if (!vst[j])
 45             {
 46                 if (t == 0) break;
 47                 t--;
 48             }
 49             s[i] = j;
 50             vst[j] = 1;
 51             k %= fac[n-i-1];
 52     }
 53 }
 54 
 55 int dir[4][2]={0,1,0,-1,1,0,-1,0};
 56 
 57 int vis[370000];
 58 char path[370000];
 59 char ansstr[370000];
 60 
 61 struct node
 62 {
 63     int state,pdir;
 64     int x,y;
 65 }Q[370000];
 66 int front,tail;
 67 
 68 int finish_state;
 69 
 70 int bfs(node start)
 71 {
 72     memset(vis,-1,sizeof(vis));
 73     vis[start.state] = 100000000;
 74     front = tail = 0;
 75     Q[tail++] = start;
 76     while (front < tail)
 77     {
 78         node now = Q[front++];
 79         if (now.state == finish_state)
 80             return finish_state;
 81         for(int i=0; i<4; i++)
 82         {
 83             node tmp = now;
 84             if (tmp.pdir != i)
 85             {
 86                 int s[10];
 87                 invKT(9,tmp.state,s);
 88 
 89                 int x = now.x, y = now.y;
 90                 int nx = x+dir[i][0],ny = y+dir[i][1];
 91                 if (0<=nx&&nx<=2 && 0<=ny&&ny<=2)
 92                 {
 93                     swap(s[x*3+y],s[nx*3+ny]);
 94                     tmp.pdir = i;
 95                     tmp.x = nx;
 96                     tmp.y = ny;
 97                     tmp.state = KT(9,s);
 98                     if (vis[tmp.state] != -1)
 99                         continue;
100                     Q[tail++] = tmp;
101                     vis[tmp.state] = now.state;
102                     char c;
103                     switch(i)
104                     {
105                     case 0:c = 'r'; break;
106                     case 1:c = 'l'; break;
107                     case 2:c = 'd'; break;
108                     case 3:c = 'u'; break;
109                     }
110                     path[tmp.state] = c;
111                 }
112             }
113         }
114     }
115     return -1;
116 }
117 
118 int main(void)
119 {
120     char str[50],tstr[3];
121     int tmps[9]={1,2,3,4,5,6,7,8,9};
122     while (~scanf("%s",tstr))
123     {
124         str[0] = tstr[0];
125         for(int i=1; i<9; i++)
126         {
127             scanf("%s",tstr);
128             str[i*2] = tstr[0];
129         }
130         int s[9];
131         int x;
132         for(int i=0; i<9; i++)
133         {
134             if ('1'<=str[i*2] && str[i*2]<='9')
135                 s[i] = str[i*2]-'0';
136             else
137             {
138                 s[i] = 9;
139                 x = i;
140             }
141         }
142         // 输入略恶心，最后用s[]记录矩阵
143         node start;
144         start.pdir = -1;
145         start.x = x/3;
146         start.y = x%3;
147         start.state = KT(9,s);
148 
149         finish_state = KT(9,tmps);
150         int ans = bfs(start);
151 
152         if (ans < 0)
153         {
154             printf("unsolvable\n");
155         }
156         else
157         {
158             int sum = 0;
159             while (vis[finish_state] != -1)
160             {
161                 if (vis[finish_state] == 100000000)
162                     break;
163                 ansstr[sum++] = path[finish_state];
164                 finish_state = vis[finish_state];
165             }
166             ansstr[sum] = 0;
167 
168             for(int i = sum-1; i>=0; i--)
169                 printf("%c",ansstr[i]);
170             printf("\n");
171         }
172     }
173     return 0;
174 }

 

转载于:https://www.cnblogs.com/toufu/p/3704858.html