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. 

4
10
20
30
04
5
2
3
4
3
4

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;

}

Hat’s Words

a
ahat
hat
hatword
hziee
word

ahat
hatword

题目分析 :

        字典树的题目, 这个题的方法比较暴力,  在输入的时候将每个单词都加入字典树,  并用数组将所有的串都存起来来.

在输入完成后,  对每个单词进行拆分, 也就是暴力枚举单词不同位置的前后部分, 看在字典树中是否存在, 存在即输出.

不过貌似数据比较弱, 说是按字典顺序输出, 但其实他的输入本就是按字典顺序输入的, 所以将排序也面了 , 呵呵.

    另外,其实这题用STL 做更简单, 当然追求效率除外.

trie 代码:

 /*

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   : 1247

*/

#include <iostream>

#include <algorithm>

#include <string>

using namespace std;

typedef struct dictor DIC;

DIC *root = NULL;

struct dictor {

       dictor (){ exist = false; memset ( child , 0 , sizeof ( child ) ); }

       void insert ( char *ins );

       bool find ( const char *ins );

private:

       DIC *child[26];

       bool exist; 

};

void dictor::insert ( char *ins )

{

            DIC *cur = root,*now;

            int len = strlen ( ins );

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

            {

                  if ( cur->child[ ins[i] - 'a' ] != NULL ) 

                  {

                       cur = cur->child[ ins[i] - 'a' ];

                  }

                  else

                  {

                       now = new DIC;

                       cur->child[ ins[i] - 'a' ] = now;

                       cur = now;

                  }

            } 

            cur->exist = true;

}

bool dictor::find ( const char *ins )

{

            DIC *cur = root;

            int len = strlen ( ins );

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

            {

                 if ( cur->child[ ins[i] - 'a' ] != NULL )

                      cur = cur->child[ ins[i] - 'a' ];  

                 else

                      return false; 

            } 

            return cur->exist;

}

char words[50050][100];

char s1[100],s2[100];

DIC dict;

int main ()

{

    root = &dict;

    int n = 0;

    while ( scanf ( "%s",words[n] ) != EOF )

    {

            dict.insert ( words[n++] );

    }

    for ( int i = 0; i < n; ++ i )

    {

          int len = strlen ( words[i] );

          for ( int j = 1; j < len; ++ j )

          {

                memset ( s1, 0, sizeof ( s1 ) );

                memset ( s2, 0, sizeof ( s2 ) );

                strncpy ( s1, words[i], j );

                strcpy ( s2, words[i]+j );

                if ( dict.find ( s1 ) && dict.find ( s2 ) )

                {

                     printf ( "%s\n", words[i] );

                     break; 

                }

          }  

    }  

    //system ( "pause" );

    return 0;

}

STL  代码 :

 /*

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

          http://www.cnblog.com/MiYu

Author By : MiYu

Test      : 2

Program   : 1247

*/

#include <iostream>

#include <string>

#include <map>

using namespace std;

map < string , int > mp;

string str[50005];

int main ()

{

    int n = 0;

    while ( cin >> str[n] ) mp[ str[n++] ] = 1;

    for ( int i = 0; i < n; ++ i )

    {

          unsigned len = str[i].size ();

          for ( unsigned j = 1; j < len; ++ j )

          {

               string s1 ( str[i], 0, j );

               string s2 ( str[i], j );

               if ( mp[s1] == 1 && mp[s2] == 1 )

               {

                    cout << str[i] << endl; 

                    break;

               }

          } 

    }

    return 0;

}

统计难题

banana
band
bee
absolute
acm

ba
b
band
abc

2
3
1
0