题目链接：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.
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.
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
题目大意：输出出现次数最多的字符串。
解题思路：每次建一条树就在结尾的时候标记一下次数，找到最大的最后输出对应的字符串就可以了。
详见代码。
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 ",cc); 71 } 72 return 0; 73 }