POJ 2488 A Knight's Journey

题目链接： http://poj.org/problem?id=2488

被字典序弄疯了，干脆枚举所有情况，然后找个字典序最小的吧。。不过代码还是很短的。

 1 #include <stdio.h>

 2 #include <string.h>

 3 

 4 struct Point

 5 {

 6     int x, y;

 7 };

 8 

 9 int map[30][30], n, m;

10 const int dir[8][2] = {{-1, -2}, {1, -2}, {-2, -1}, {2, -1}, {-2, 1}, {2, 1}, {-1, 2}, {1, 2}};

11 bool vis[30][30];

12 char path[60], ans[60];

13 

14 void dfs(int x, int y, int cnt)

15 {

16     vis[x][y] = 1;

17     path[cnt++] = y + 'A' - 1;

18     path[cnt++] = x + '0';

19     if(cnt / 2 == n*m)

20     {

21         path[cnt] = 0;

22         if(ans[0] == 0 || strcmp(path, ans) < 0)

23             strcpy(ans, path);

24     }

25     for(int i = 0; i < 8; i++)

26     {

27         int nx = x + dir[i][0];

28         int ny = y + dir[i][1];

29         if(nx > 0 && nx <= n && ny > 0 && ny <= m && !vis[nx][ny])

30             dfs(nx, ny, cnt);

31     }

32     vis[x][y] = 0;

33 }

34 

35 int main()

36 {

37     int t;

38     scanf("%d", &t);

39     for(int i = 1; i <= t; i++)

40     {

41         scanf("%d %d", &n, &m);

42         printf("Scenario #%d:\n", i);

43         memset(vis, 0, sizeof(vis));

44         ans[0] = 0;

45         dfs(1, 1, 0);

46         if(ans[0] == 0)

47             printf("impossible\n\n");

48         else

49             printf("%s\n\n", ans);

50     }

51     return 0;

52 }

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