Hat’s Words (分成两个字符串考虑)

Posted 2022-04-24 -ackerman

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1247

 

思路：

把字符串分成两个部分，看这两个部分是不是也在字典树上。如果在，就输出这个字符串。 

 

这题我只想说静态内存真是容易爆！

 1 #include <stdio.h>
 2 #include <string.h>
 3 #include <string>
 4 #include <iostream>
 5 #include <stdlib.h>
 6 #include <algorithm>
 7 
 8 using namespace std;
 9 const int MAX_NODE = 100000 + 10;
10 const int CHARSET = 26;
11 
12 int trie[MAX_NODE][CHARSET]=0;
13 int color[MAX_NODE]=0;
14 int vis[MAX_NODE];
15 char s[50001][300];
16 int k = 1;
17 
18 void insert(char *w)
19 
20     int len = strlen(w);
21     int p = 0;
22     for (int i=0;i<len;i++)
23     
24         int c = w[i]-‘a‘;
25         if (!trie[p][c])
26         
27             trie[p][c] = k;
28             k++;
29         
30         p = trie[p][c];
31         vis[p]++;
32     
33     color[p] = 1;
34 
35 
36 
37 int search(char *s)
38 
39     int len = strlen(s);
40     int p = 0;
41     for (int i=0;i<len;i++)
42     
43         int c = s[i]-‘a‘;
44         if (!trie[p][c])
45             return 0;
46         p = trie[p][c];
47     
48     return color[p] == 1;
49 
50 
51 
52 int main()
53 
54 #ifndef ONLINE_JUDGE
55     freopen("../in.txt","r",stdin);
56 #endif
57     int t = 0;
58     while (scanf("%s",s[t])!=EOF)
59     
60         insert(s[t]);
61         t++;
62     
63     for (int i=0;i<t;i++)
64     
65         for (int j=1;j<strlen(s[i]);j++)
66         
67             char str1[1000]=‘\0‘;
68             char str2[1000]=‘\0‘;
69             strncpy(str1,s[i],j);
70             strncpy(str2,s[i]+j,strlen(s[i])-j);
71             if(search(str1) && search(str2))
72             
73                 cout << s[i] << endl;
74                 break;
75             
76         
77     
78     return 0;
79