• ZOJ 1217 eight


    八数码,双广、A*都超时了(可能是写得不好),IDA*通过了;

    在POJ上是16MS(和双广一样,A* 60MS左右),HDOJ上是800MS左右(双广是500MS),ZOJ上只有IDA*没超时(2480MS)。

      1 # include <stdio.h>
      2 # include <math.h>
      3 # include <string.h>
      4 
      5 # define MIN(x, y) ((x)<(y) ? (x):(y))
      6 
      7 # define N 9
      8 # define DIR_N 4
      9 # define INF 0x7fffffff
     10 # define MAX_DEPTH 1000
     11 # define SIZE 3
     12 
     13 char solved;
     14 char start[N], goal[N];
     15 char cur[N];
     16 
     17 int bound;
     18 const char md[] = {'u', 'l', 'r', 'd'};
     19 const short int dir[][2] = {{-1,0}, {0,-1}, {0,1}, {1,0}};
     20 char path[MAX_DEPTH];
     21 
     22 void solve(void);
     23 char read(char *s);
     24 int IDAstar(void);
     25 char bool_inv(char *s);
     26 int heu(void);
     27 int dfs(int zero_pos, int depth, char direction);
     28 
     29 int main()
     30 {
     31     int i;
     32     
     33     for (i = 0; i < N-1; ++i)
     34         goal[i] = i + 1;
     35     goal[N-1] = 0;
     36     while (read(start))
     37     {
     38         memcpy(cur, start, N);
     39         solve();
     40     }
     41     
     42     return 0;
     43 }
     44 
     45 void solve()
     46 {
     47     int i, depth;
     48 
     49     if (bool_inv(start) != bool_inv(goal)) puts("unsolvable");
     50     else
     51     {
     52         memset(path, -1, sizeof(path));
     53         depth = IDAstar();
     54         for (i = 0; path[i]!=-1; ++i)
     55         {
     56             putchar(md[path[i]]);
     57         }
     58         putchar('\n');
     59     }
     60 }
     61 
     62 char read(char *s)
     63 {
     64     char c[5];
     65     int i;
     66 
     67     if (EOF == scanf("%s", c)) return 0;
     68     else s[0] = (c[0]=='x' ? 0:c[0]-'0');
     69     for (i = 1; i < N; ++i)
     70     {
     71         scanf("%s", c);
     72         s[i] = (c[0]=='x' ? 0:c[0]-'0');
     73     }
     74     
     75     return 1;
     76 }
     77 
     78 int IDAstar(void)
     79 {
     80     int i;
     81 
     82     solved = 0;
     83     bound = heu();
     84 
     85     for (i = 0; start[i] && i < N; ++i) ;
     86 
     87     while (!solved && bound<100)
     88     {
     89         bound = dfs(i, 0, -10);
     90     }
     91 
     92     return bound;
     93 }
     94 
     95 int dfs(int pos, int depth, char d)
     96 {
     97     int h, i, nx, ny, npos, nbound, t;
     98 
     99     h = heu();
    100 
    101     if (depth+h > bound) return depth+h;
    102     if (h == 0)
    103     {
    104        /* printf("depth = %d.\n", depth); */
    105         solved = 1;
    106         return depth;
    107     }
    108 
    109     nbound = INF;
    110     for (i = 0; i < DIR_N; ++i)
    111     {
    112         if (i+d == DIR_N-1) continue;
    113         nx = pos/SIZE + dir[i][0];
    114         ny = pos%SIZE + dir[i][1];
    115         if (0<=nx && nx<SIZE && 0<=ny && ny<SIZE)
    116         {
    117             path[depth] = i;
    118             npos = nx*SIZE + ny;
    119             cur[pos] = cur[npos]; cur[npos] = 0;     /* pos -> npos */
    120             t = dfs(npos, depth+1, i);
    121             if (solved) return t;
    122             nbound = MIN(nbound, t);
    123             cur[npos] = cur[pos]; cur[pos] = 0;
    124         }
    125     }
    126     return nbound;
    127 }
    128 
    129 int heu(void)            /* return heu(cur_state) */
    130 {
    131     char ch;
    132     int i, j, ret;
    133 
    134     ret = 0;
    135     for (i = 0; i < N; ++i)
    136     {
    137         ch = goal[i];
    138         if (ch == 0) continue;
    139         for (j = 0; j < N; ++j)
    140         {
    141             if (ch == cur[j])
    142             {
    143                 ret = ret + abs(i/SIZE-j/SIZE) + abs(i%SIZE-j%SIZE);
    144             }
    145         }
    146     }
    147 
    148     return ret;
    149 }
    150 
    151 char bool_inv(char *s)
    152 {
    153     char ch, ret;
    154     int i, j;
    155 
    156     ret = 0;
    157     for (i = 0; i < N-1; ++i)
    158     {
    159         if ((ch = s[i]) == 0) continue;
    160         for (j = i+1; j < N; ++j)
    161         {
    162             if (s[j] && s[j]<ch) ret = 1 - ret;
    163         }
    164     }
    165 
    166     return ret;
    167 }
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  • 原文地址:https://www.cnblogs.com/JMDWQ/p/2519734.html
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