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HDOJ 1800 HDU 1800 Phone List ACM 1800 IN HDU

Posted on 2010-08-25 11:54 MiYu 阅读(582) 评论(0)  编辑 收藏 引用 所属分类: ACM ( 串 ) 、ACM ( 数据结构 ) 、ACM ( 字典树( TRIE ) )

MiYu原创, 转帖请注明 : 转载自 ______________白白の屋    

 

题目地址:

      http://acm.hdu.edu.cn/showproblem.php?pid=1800 

题目描述:

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
 

题目分析:

 

相对来说这题用字典树的效率比较高些.....遗憾的是我现在只会做一些字典树的 水题....对字典树的理解现在还不是很好......所以,呵呵

这题我就直接筛选过了........

 

代码如下:

 /*

MiYu原创, 转帖请注明 : 转载自 ______________白白の屋

          http://www.cnblog.com/MiYu

Author By : MiYu

Test      : 1

Program   : 1800

*/


#include <iostream>

#include <algorithm>

using namespace std;

int num[3002];

bool hash[3002];

bool cmp ( const int &a, const  int &b )

{

     return a > b; 

}

int main ()

{

    int N;

    while ( scanf ( "%d", &N ) != EOF )

    {

           memset ( num, 0, 12008 );

           memset ( hash, 0, 3002 );

           for ( int i = 0; i != N; ++ i )

                 scanf ( "%d",num + i );

           sort ( num, num + N, cmp );

           int nCount = 0;

           for ( int i = 0; i < N; ++ i )

           {

                 if ( !hash[i] )

                 {

                      nCount ++;

                      int j = i ,k = j+1;

                      hash[j] = true;

                      while ( k < N ) 

                      {

                            if ( !hash[k] && num[k] < num[j] )

                            {

                                 hash[k] = true;

                                 j = k;

                            }

                            k++;

                      } 

                 }

           } 

           printf ( "%d\n",nCount );

    }

    return 0;

}


 

 

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