Flying to the Mars

D - Flying to the Mars

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 1800

Description

In the year 8888, the Earth is ruled by the PPF Empire . As the population growing , PPF needs to find more land for the newborns . Finally , PPF decides to attack Kscinow who ruling the Mars . Here the problem comes! How can the soldiers reach the Mars ? PPF convokes his soldiers and asks for their suggestions . “Rush … ” one soldier answers. “Shut up ! Do I have to remind you that there isn’t any road to the Mars from here!” PPF replies. “Fly !” another answers. PPF smiles :“Clever guy ! Although we haven’t got wings , I can buy some magic broomsticks from HARRY POTTER to help you .” Now , it’s time to learn to fly on a broomstick ! we assume that one soldier has one level number indicating his degree. The soldier who has a higher level could teach the lower , that is to say the former’s level > the latter’s . But the lower can’t teach the higher. One soldier can have only one teacher at most , certainly , having no teacher is also legal. Similarly one soldier can have only one student at most while having no student is also possible. Teacher can teach his student on the same broomstick .Certainly , all the soldier must have practiced on the broomstick before they fly to the Mars! Magic broomstick is expensive !So , can you help PPF to calculate the minimum number of the broomstick needed . 
For example : 
There are 5 soldiers (A B C D E)with level numbers : 2 4 5 6 4; 
One method : 
C could teach B; B could teach A; So , A B C are eligible to study on the same broomstick. 
D could teach E;So D E are eligible to study on the same broomstick; 
Using this method , we need 2 broomsticks. 
Another method: 
D could teach A; So A D are eligible to study on the same broomstick. 
C could teach B; So B C are eligible to study on the same broomstick. 
E with no teacher or student are eligible to study on one broomstick. 
Using the method ,we need 3 broomsticks. 
……

After checking up all possible method, we found that 2 is the minimum number of broomsticks needed.

 

Input

Input file contains multiple test cases. 
In a test case,the first line contains a single positive number N indicating the number of soldiers.(0<=N<=3000) 
Next N lines :There is only one nonnegative integer on each line , indicating the level number for each soldier.( less than 30 digits); 

 

Output

For each case, output the minimum number of broomsticks on a single line.

 

Sample Input

4
10
20
30
04
5
2
3
4
3
4

 

Sample Output

1
2

 

思路：统计输入的数的众数，小于众数的数是众数的学生，大于众数的数是众数的老师，故而众数就是答案。

输入的数不多于30digits，64位整数存放不下，32位更不行，但是很多人用32位或者64位做出来了，原因应该不是数据太弱，而是就算溢出也不影响众数的统计。以下程序用的是大数。另一程序用的是字符串哈希，让字符串映射一个整数。

AC Code：

#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <map>
#include <string>
#include <cstdio>
#include <cmath>
#include <cstring>
 
using namespace std;
 
const int maxn = ;
struct Level
{
    char l[];
    int len;
}a[maxn];
int n;
 
bool cmp(const Level& x, const Level& y)
{
    if(x.len != y.len) return x.len < y.len;
    for(int i = ; i >=  - x.len; i--)
    {
        if(x.l[i] > y.l[i]) return false;
        return true;
    }
}
 
int main()
{
    char c;
    while(scanf("%d%c", &n, &c) != EOF)
    {
        for(int i = ; i < n; i++)
        {
            memset(a[i].l, '', sizeof(a[i].l));
            a[i].l[] = '\0';
            a[i].len = ;
            for(int j = ; scanf("%c", &c) && c != '\n';)
            {
                if(j ==  && c == '') continue;
                a[i].l[j] = c;
                a[i].len++;
                j--;
            }
        }
        sort(a, a + n, cmp);
        int max = -;
        for(int i = ; i < n; )
        {
            int j;
            for(j = i + ; j < n; j++)
            {
                if(strcmp(a[i].l, a[j].l)) break;
            }
            if(max < j - i) max = j - i;
            i = j;
        }
        printf("%d\n", max);
    }
    return ;
}

方法二：

 #include <iostream>
 #include <algorithm>
 #include <vector>
 #include <queue>
 #include <map>
 #include <string>
 #include <cstdio>
 #include <cmath>
 #include <cstring>
 
 using namespace std;
 
 const int MOD = ;
 const int MAXN = ;
 const int LEN = ;
 const int seed[] = {, };
 int level[MAXN];
 char s[LEN];
 
 int Hash()
 {
     int res = ;
     int i;
     for(i = ; s[i] == ''; i++){}
     for(; s[i] != '\0'; i++)
     {
         res += ((res * seed[s[i]&] + s[i]) % MOD);
     }
     return res;
 }
 
 int main()
 {
     int n;
     while(scanf("%d", &n) != EOF)
     {
         for(int i = ; i < n; i++)
         {
             scanf("%s", s);
             level[i] = Hash();
         }
         sort(level, level + n);
         int max = ;
         for(int i = ; i < n;)
         {
             int j;
             for(j = i + ; j < n && level[j] == level[i]; j++){}
             if(max < j - i)
                 max = j - i;
             i = j;
         }
         printf("%d\n", max);
     }
     return ;
 }