【HDU1198】Farm Irrigation（回溯+记忆化搜索）

数据流小，深搜即可。有些暴力。看其他人的题解用二维转换成一维做的并查集很巧妙，马上去研究一下！！

 

 1 #include <iostream>

 2 #include <cstring>

 3 #include <cstdlib>

 4 #include <cstdio>

 5 #include <cmath>

 6 #include <cctype>

 7 #include <algorithm>

 8 #include <numeric>

 9 #include <string>

10 #include <limits.h>

11 #include <map>

12 using namespace std;

13 

14 int m, n, ans = 0;

15 char Map[55][55];

16 bool vis[55][55];

17 map<char, string> check;

18 

19 void dfs (int x, int y) {

20 

21     vis[x][y] = 1;

22 

23     if (check[Map[x][y]].find("U") != -1 && check[Map[x - 1][y]].find("D") != -1) {

24         if (!vis[x - 1][y]) dfs (x - 1, y);

25     }

26 

27     if (check[Map[x][y]].find("D") != -1 && check[Map[x + 1][y]].find("U") != -1) {

28         if (!vis[x + 1][y]) dfs (x + 1, y);

29     }

30 

31     if (check[Map[x][y]].find("L") != -1 && check[Map[x][y - 1]].find("R") != -1) {

32         if (!vis[x][y - 1]) dfs (x, y - 1);

33     }

34 

35     if (check[Map[x][y]].find("R") != -1 && check[Map[x][y + 1]].find("L") != -1) {

36         if (!vis[x][y + 1]) dfs (x, y + 1);

37     }

38 }

39 

40 int main () {

41 

42     check['A'] = "UL";  check['B'] = "UR";  check['C'] = "DL";

43     check['D'] = "DR";  check['E'] = "UD";  check['F'] = "LR";

44     check['G'] = "ULR"; check['H'] = "UDL"; check['I'] = "DLR";

45     check['J'] = "UDR"; check['K'] = "UDLR";

46 

47     while (~scanf("%d%d", &m, &n)) {

48         if (m < 0 && n < 0) {

49             break;

50         }

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

52         memset (Map, 0, sizeof(Map));

53         for (int i = 0; i < m ; ++ i) {

54             for (int j = 0 ; j < n; ++ j) {

55                 cin >> Map[i][j];

56             }

57         }

58         ans = 0;

59         for (int i = 0; i < m; ++ i) {

60             for (int j = 0; j < n; ++j) {

61                 if (vis[i][j] != 1) {

62                     dfs (i, j);

63                     ans ++;

64                 }

65             }

66         }

67 

68         cout << ans << endl;

69     }

70     return 0;

71 }