hdu 1800 Flying to the Mars

Flying to the Mars

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 19423 Accepted Submission(s): 6254

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

题意：给你n个士兵，现在每个士兵都有代表自己骑扫帚技术水平高低的数字，要求：只能是水平高的教导水平低的，且没人最多有一个徒弟，同样每个人最多有一个师傅，师傅可以教导徒弟在同一个扫帚上面，问最少需要多少扫帚。

思路：简单贪心。  我们可以对其进行排序，然后每次找水平闭目前这个士兵高的，看看能找到多少组。

求上升序列的。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int main()
{
int a[3001], vis[3001];
int n;
while(~scanf("%d",&n))
{
for(int i = 0; i < n; i++)
{
scanf("%d",&a[i]);
}
sort(a, a + n);
int m = 0, sum = 0;
memset(vis, 0, sizeof(vis));
for(int i = 0; i < n; i++)
{
m = a[i];
if(vis[i]!=1)
{
for(int j = i + 1; j < n; j++)
{
if(vis[j]!=1&&a[j] > m)
{
m = a[j];
vis[j] = 1;
}
}
sum++;
}

}
printf("%d\n",sum);
}

return 0;
}