HDU 1800 Flying to the Mars (贪心)

Flying to the Mars

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8651    Accepted Submission(s): 2823

Problem 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

410203004523434

 

Sample Output

12

 题目大意：就是一群人去火星,坐扫把去。但每个人都要练习扫把（题目中这样说的:all the soldier must have practiced on the broomstick before they fly to the Mars!），每个人都有等级，等级高的人可以教等级低的人练习使用扫把。一个人最多教一个人，一个人最多有一个老师。（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）。如果一个人没有老师，也没有学生，单独使用一个扫把。问题：因为扫把贵，要求再输入这些人，用最小的扫把教会这么多人。（所有人等级小于30）

思路：对于每一个人。要不当老师，要不当学生。要不单独使用。所以给所有等级排序。相同等级且重复最多的个数。就是最少扫把数。因为对于这些人，其他人要不是老师，要不是学生。都可以共用一把扫把。然而，他们等级相等，只能自己教自己。

一次AC。。。

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int main(){    int a[3010];    int n,i;    int max,count;    while(scanf("%d",&n)!=EOF)    {        max=1;        count=1;        for(i=0;i<n;i++)        scanf("%d",&a[i]);        sort(a,a+n);        for(i=1;i<n;i++)        {            if(a[i-1]==a[i])        //找到等级相等的的最多士兵个数，即最少需要的扫把数目            {                count++;                if(count>max)                max=count;                continue;            }            count=1;        }        printf("%d\n",max);    }    return 0;}