UVa 1419 - Ugly Windows (DFS + 暴力)

题意

找出有几个在顶端的窗口。需要满足以下条件

1. 窗口里面只有'.'
2. 窗口边缘是完整的。

思路

在浏览器标签上呆了一个星期的题目=。= 
今天标签实在是放不下了，就去A这题，空一点位置╮(╯▽╰)╭

1. 找出完整的窗口。
2. 判断这个窗口里面是不是全部是.

如何找出完整的窗口？ 
碰到一个字母就DFS，如果从下往上走到起点，说明这个窗口是完整的。因此在起点的时候不能往下走。

如何判断2？ 
枚举出当前窗口的边界，for一下。

代码

1. #include
2. #include
3. #include
4. #include
5. #include
6. #include
7. #include
8. #include
9. #include
10. #include
11. #include
12. #include
13. #include
14. #include
15. #define LL long long
16. #define ULL unsigned long long
17. #define SZ(x) (int)x.size()
18. #define Lowbit(x) ((x) & (-x))
19. #define MP(a, b) make_pair(a, b)
20. #define MS(arr, num) memset(arr, num, sizeof(arr))
21. #define PB push_back
22. #define F first
23. #define S second
24. #define ROP freopen("input.txt", "r", stdin);
25. #define MID(a, b) (a + ((b - a) >> 1))
26. #define LC rt << 1, l, mid
27. #define RC rt << 1|1, mid + 1, r
28. #define LRT rt << 1
29. #define RRT rt << 1|1
30. #define BitCount(x) __builtin_popcount(x)
31. #define BitCountll(x) __builtin_popcountll(x)
32. #define LeftPos(x) 32 - __builtin_clz(x) - 1
33. #define LeftPosll(x) 64 - __builtin_clzll(x) - 1
34. const double PI = acos(-1.0);
35. const int INF = 0x3f3f3f3f;
36. using namespace std;
37. const double eps = 1e-8;
38. const int MAXN = 100 + 10;
39. const int MOD = 1000007;
40. const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} }; //0123，上下左右
42. typedef pair<int, int> pii;
43. typedef vector::iterator viti;
44. typedef vector::iterator vitii;
46. char mp[MAXN][MAXN];
47. int vis[MAXN][MAXN], n, m;
48. pii tar;
49. vector ans;
51. bool Check(char curWindow)
52. {
53. int i, j;
54. for (i = tar.F, j = tar.S; mp[i][j] == curWindow; j++);
55. int len = j - 2;
56. for (i = tar.F, j = tar.S; mp[i][j] == curWindow; i++);
57. int wide = i - 2;
58. for (i = tar.F + 1; i <= wide; i++)
59. for (j = tar.S + 1; j <= len; j++)
60. if (mp[i][j] != '.') return false;
61. return true;
62. }
64. void DFS(int x, int y, char curWindow)
65. {
66. if (mp[x][y] == '.' || mp[x][y] != curWindow) return;
67. vis[x][y] = 1;
68. for (int i = 0; i < 4; i++)
69. {
70. if (i == 1 && x == tar.F && y == tar.S) continue; //一开始的时候不能往下走
71. int xx = x + dir[i][0], yy = y + dir[i][1];
72. if (i == 0 && xx == tar.F && yy == tar.S) //如果往上走能到达起点
73. {
74. if (Check(curWindow))
75. ans.PB(curWindow);
76. }
77. if (xx >= 1 && xx <= n && yy >= 1 && yy <= m && !vis[xx][yy])
78. DFS(xx, yy, curWindow);
79. }
80. }
82. int main()
83. {
84. //ROP;
85. int i, j;
86. while (scanf("%d%d", &n, &m), n + m)
87. {
88. ans.clear();
89. MS(vis, 0);
90. for (i = 1; i <= n; i++) scanf("%s", mp[i] + 1);
91. for (i = 1; i <= n; i++)
92. for (j = 1; j <= m; j++)
93. if (!vis[i][j] && mp[i][j] != '.')
94. {
95. tar.F = i, tar.S = j;
96. DFS(i, j, mp[i][j]);
98. }
99. sort(ans.begin(), ans.end());
100. for (i = 0; i < SZ(ans); i++) putchar(ans[i]);
101. puts("");
102. }
103. return 0;
104. }