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
4
10
20
30
04
5
2
3
4
3
4
Sample Output
1
2
这道题的思路还是挺简单的，就是寻找所有数中出现次数最大的那一个数，但是上面说最长达到了30位数，所以用普通的排序做不了，这时候可以用map容器来做，map容器的内部是一个红黑树，我们是在对它的叶节点进行操作，一共有两个数，第二个数是作为计数用的。
1 #include <map>
2 using namespace std;
3
4 int main(){
5 int number;
6 int i;
7 int level;
8 int max;
9
10 while(scanf("%d",&number)!=EOF){
11 max=0;
12
13 map<int,int> mp;
14 for(i=0;i<number;i++){
15 scanf("%d",&level);
16 mp[level]++;
17
18 if(mp[level]>max)
19 max=mp[level];
20 }
21
22 printf("%d\n",max);
23 }
24
25 return 0;
26 }