hdu 1004 Let the Balloon Rise（字典樹）

題目連接：http://acm.hdu.edu.cn/showproblem.php?pid=1004

Let the Balloon Rise

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 90644    Accepted Submission(s): 34459

Problem Description

Contest time again! How excited it is to see balloons floating around. But to tell you a secret, the judges' favorite time is guessing the most popular problem. When the contest is over, they will count the balloons of each color and find the result.

This year, they decide to leave this lovely job to you.

 

 

Input

Input contains multiple test cases. Each test case starts with a number N (0 < N <= 1000) -- the total number of balloons distributed. The next N lines contain one color each. The color of a balloon is a string of up to 15 lower-case letters.

A test case with N = 0 terminates the input and this test case is not to be processed.

 

 

Output

For each case, print the color of balloon for the most popular problem on a single line. It is guaranteed that there is a unique solution for each test case.

 

 

Sample Input

5

green

red

blue

red

red

3

pink

orange

pink

0

 

Sample Output

red

pink

 

Author

WU, Jiazhi

 

題目大意：輸出出現次數最多的字符串。

解題思路：每次建一條樹就在結尾的時候標記一下次數，找到最大的最後輸出對應的字符串就能夠了。

 

詳見代碼。

 

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 
 5 using namespace std;
 6 
 7 struct node
 8 {
 9     node *next[26];
10     int Count;
11     node()
12     {
13         for (int i=0; i<26; i++)
14             next[i]=NULL;
15         Count=0;
16     }
17 };
18 
19 char cc[5000];
20 node *p,*root=new node();
21 void insert(char *s)
22 {
23     p=root;
24     for (int i=0; s[i]; i++)
25     {
26         int k=s[i]-'a';
27         if (p->next[k]==NULL)
28             p->next[k]=new node();
29         p=p->next[k];
30     }
31     p->Count++;
32 }
33 
34 int Search(char *s)
35 {
36     p=root;
37     for (int i=0; s[i]; i++)
38     {
39         int k=s[i]-'a';
40         if (p->next[k]==NULL)
41             return 0;
42         p=p->next[k];
43     }
44     //cout<<p->Count<<" "<<"3333333333"<<endl;
45     return p->Count;
46 }
47 
48 int main()
49 {
50     int t;
51     char ch[1010];
52     int Max;
53     while (~scanf("%d",&t))
54     {
55         Max=0;
56         if (t==0)
57             break;
58         while (t--)
59         {
60             scanf("%s",ch);
61             //gets(ch);
62             insert(ch);
63             int ans=Search(ch);
64             if (ans>Max)
65             {
66                 Max=ans;
67                 strcpy(cc,ch);
68             }
69         }
70         printf ("%s\n",cc);
71     }
72     return 0;
73 }