Let the Balloon Rise

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

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

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

Source

ZJCPC2004

AC1字典树：

#include<cstdio>
#include<cmath>
#include<cstring>
#include<cstdlib>
#include<stack>
#include<queue>
#include<vector>
#include<algorithm>
using namespace std;
#define INF 0x7fffffff;
struct node
{
    int x;
    node *next[26];
    node()
    {
        x=0;
        memset(next,0,sizeof(next));
    }
};
node *root=NULL;

void insert(char *s)
{
    node *p=root;
    int i,k;
    for(i=0;i<strlen(s);i++)
    {
      k=s[i]-'a';
      if(p->next[k]==NULL)
          p->next[k]=new node;
      else
          p->next[k]->x++;
      p=p->next[k];
    }
}

int search(char *s)
{
    node *p=root;
    int i,k;
    for(i=0;i<strlen(s);i++)
    {
        k=s[i]-'a';
        if(p->next[k]==NULL)
         return 0;
          p=p->next[k];
    }
  return p->x;
}

char str[1010][30];
int main()
{

    int n;
    root=new node;
    while(scanf("%d",&n),n)
    {
        int i=0;
        memset(str,0,sizeof(str));
        for(i=0;i<n;i++)
        {
          scanf("%s",str[i]);
          insert(str[i]);
        }

       int ma=-INF;
       int j=0;
       for(i=0;i<n;i++)
       {
        int k=search(str[i]);
           if(k>ma)
           {
            ma=k;
         j=i;
        }

       }
       printf("%s\n",str[j]);
    }
    return 0;
}

AC2 模拟：

#include<cstdio>
#include<string.h>
#include<stdlib.h>
const int max= 1000+10;
char ba[max][20];
int b[max];

int main()
{
    int N;
    while(scanf("%d",&N),N)
    {
        int i,j;

        memset(ba,0,sizeof(ba));
        getchar();
        for(i=0;i<N;i++)
          scanf("%s",ba[i]);

         memset(b,0,sizeof(b));
          for(i=0;i<N;i++)
              for(j=i+1;j<N;j++)
                 if(strcmp(ba[i],ba[j])==0)
                   b[i]++;

            int m=0,t;
            for(j=0;j<N;j++)
                if(m<=b[j])
                  m=b[j];

            for(i=0;i<N;i++)
                  if(b[i]==m)
                        {
                      printf("%s\n",ba[i]);
                        break;
                    }

    }
    system("pause");
    return 0;
}