zoukankan      html  css  js  c++  java
  • 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 }
  • 相关阅读:
    三元判断大小奇偶
    键盘输入 循环 for if-else while 等循环
    三元 导包
    进制
    符号等
    第二模块 4.1 模块的种类和介绍
    第二模块 练习题
    第二模块 3.15 迭代器
    第二模块 3.14 通过生成器实现并发编程
    第二模块 3.13 生成器
  • 原文地址:https://www.cnblogs.com/z-712/p/7307327.html
Copyright © 2011-2022 走看看