http://acm.hdu.edu.cn/showproblem.php?pid=1800
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.
4 10 20 30 04 5 2 3 4 3 4
1 2
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int a[3002];
bool cmp(int a,int b)
{
return a>b;
}
int main()
{
int n,i,j,tot,k;
while(~scanf("%d",&n))
{
k=1;
tot=0;
for(i=0;i<n;i++)
scanf("%d",&a[i]);
sort(a,a+n,cmp);
for(i=1;i<n;i++)
{
if(a[i]==a[i-1])
k++;
else
{
if(tot<k)
{
tot=k;
}
k=1;
}
}
if(k>tot)
tot=k;
printf("%d\n",tot);
}
return 0;
}
杭电 1800 Flying to the Mars（贪心）
原文：http://www.cnblogs.com/lytwajue/p/6714076.html