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  • hdoj 1004 Let the Balloon Rise

    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
     
     1 #include <iostream>
     2 #include <bits/stdc++.h>   //º¼µç²»ÈÏͬ
     3 using namespace std;
     4 typedef struct node{
     5     struct node*next[26];
     6     int value;
     7     int num;
     8     }*trie,node;
     9     trie root;
    10 
    11     void init_trie(trie&p){
    12     p=(trie)malloc(sizeof(node));
    13     for(int i=0;i<26;i++)
    14 
    15         p->next[i]=0;
    16         p->value=p->num=0;
    17     }
    18     void insert(trie p,char s[])
    19     {
    20         int k=0;
    21         while (s[k]!='')
    22         {
    23             if(!p->next[s[k]-'a'])
    24             {
    25                 trie q;
    26                 init_trie(q);
    27                 p->next[s[k]-'a']=q;
    28             }
    29             p=p->next[s[k]-'a'];
    30             p->num++;
    31             k++;
    32         }
    33         p->value=1;
    34     }
    35     int find (trie p,char s[]){
    36     int k=0;
    37     while(s[k]!=''&&p->next[s[k]-'a'])
    38     {
    39 
    40         p=p->next[s[k]-'a'];
    41         k++;
    42     }
    43     if(s[k]=='')
    44         return p->num;
    45     return 0;
    46 
    47     }
    48 
    49 int main()
    50 {  int n;
    51     while(scanf("%d",&n)!=EOF&&n)
    52     {  init_trie(root);
    53     char ch[1005][17];
    54     for(int i=0;i<n;i++)
    55     {
    56         cin>>ch[i];
    57         insert(root,ch[i]);
    58     }
    59     int max=0,tem,f;
    60     for(int i=0;i<n;i++)
    61     {
    62         tem=find(root,ch[i]);
    63         if(tem>max)
    64         {
    65             max=tem;
    66             f=i;
    67         }
    68     }
    69     cout<<ch[f]<<"
    ";
    70 
    71         }
    72         return 0;
    73 }
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  • 原文地址:https://www.cnblogs.com/z-712/p/7307327.html
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