IDA*+曼哈頓 相當快！ 通過五組數據總計用時 0.015000 s.

把宏定義中的N和SIZE改一下，就可以處理十五數碼問題了，終于修煉到第八層了；

剛開始還是使用結構體定義了狀態，寫得相當麻煩，后來發現IDA*中不像BFS和A*，只需要考慮一條路，，參考了lym（http://www.cnblogs.com/liyongmou/archive/2010/07/19/1780861.html）的（后面調試的受不了了，對照著此大牛的代碼一點點改，最后發現是更新狀態時，npos寫成了pos……）

# include <stdio.h> # include <math.h> # include <string.h> # include <time.h># define MIN(x, y) ((x)<(y) ? (x):(y))# define N 9 # define DIR_N 4 # define INF 0x7fffffff # define MAX_DEPTH 1000 # define SIZE 3char solved; char start[N], goal[N]; char cur[N];int bound; const char md[] = {'u', 'l', 'r', 'd'}; const short int dir[][2] = {{-1,0}, {0,-1}, {0,1}, {1,0}}; char path[MAX_DEPTH];void move(void); void solve(void); char read(char *s); int IDAstar(void); char bool_inv(char *s); int heu(void); int dfs(int zero_pos, int depth, char direction); void print_state(char *s);int main() {freopen("in.txt", "r", stdin);freopen("out.txt", "w", stdout);while (read(start)){read(goal);memcpy(cur, start, N);solve();}printf("time cost %.6lf s.\n", (double)clock()/CLOCKS_PER_SEC);return 0; }void solve() {int i, depth;if (bool_inv(start) != bool_inv(goal)) puts("unsolvable");else{memset(path, -1, sizeof(path));depth = IDAstar();for (i = 0; path[i]!=-1; ++i){putchar(md[path[i]]);}printf(" %d\n", i);}//move(); //打印移動過程，調試的時候用的，加上這個總共用時31ms }char read(char *s) {char c[5];int i;if (EOF == scanf("%s", c)) return 0;else s[0] = (c[0]=='x' ? 0:c[0]-'0');for (i = 1; i < N; ++i){scanf("%s", c);s[i] = (c[0]=='x' ? 0:c[0]-'0');}return 1; }int IDAstar(void) {int i;solved = 0;bound = heu();for (i = 0; start[i] && i < N; ++i) ;while (!solved && bound<100){bound = dfs(i, 0, -10);}return bound; }int dfs(int pos, int depth, char d) {int h, i, nx, ny, npos, nbound, t;h = heu();if (depth+h > bound) return depth+h;if (h == 0){/* printf("depth = %d.\n", depth); */solved = 1;return depth;}nbound = INF;for (i = 0; i < DIR_N; ++i){if (i+d == DIR_N-1) continue;nx = pos/SIZE + dir[i][0];ny = pos%SIZE + dir[i][1];if (0<=nx && nx<SIZE && 0<=ny && ny<SIZE){path[depth] = i;npos = nx*SIZE + ny;cur[pos] = cur[npos]; cur[npos] = 0; /* 后來一直出錯就是這里的原因：pos -> npos */t = dfs(npos, depth+1, i);if (solved) return t;nbound = MIN(nbound, t);cur[npos] = cur[pos]; cur[pos] = 0;}}return nbound; }int heu(void) /* return heu(cur_state) */ {char ch;int i, j, ret;ret = 0;for (i = 0; i < N; ++i){ch = goal[i];if (ch == 0) continue;for (j = 0; j < N; ++j){if (ch == cur[j]){ret = ret + abs(i/SIZE-j/SIZE) + abs(i%SIZE-j%SIZE);}}}return ret; }char bool_inv(char *s) {char ch, ret;int i, j;ret = 0;for (i = 0; i < N-1; ++i){if ((ch = s[i]) == 0) continue;for (j = i+1; j < N; ++j){if (s[j] && s[j]<ch) ret = 1 - ret;}}return ret; }void move(void) {char ncur[N];int i, pos, nx, ny, npos;memcpy(ncur, start, N);print_state(ncur);for (i = 0; start[i] && i<N; ++i) ;pos = i;for (i = 0; path[i] != -1; ++i){nx = pos/SIZE + dir[path[i]][0];ny = pos%SIZE + dir[path[i]][1];npos = nx*SIZE + ny;ncur[pos] = ncur[npos]; ncur[npos] = 0;pos = npos;print_state(ncur);} }void print_state(char *s) {int i;for (i = 0; i < N; ++i){putchar(s[i]+'0');}putchar('\n'); }

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轉載于:https://www.cnblogs.com/JMDWQ/archive/2012/05/27/2519728.html

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