【HDU1198】Farm Irrigation（回溯+记忆化搜索）

最新推荐文章于 2018-09-08 09:59:54 发布

banxuanliu0005

最新推荐文章于 2018-09-08 09:59:54 发布

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原文链接：http://www.cnblogs.com/Destiny-Gem/p/3869648.html

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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 }