hdu1800 Flying to the Mars(字典树)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1800
Flying to the Mars
![](http://acm.hdu.edu.cn/data/images/1800-1.jpg)
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.
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);
4 10 20 30 04 5 2 3 4 3 4
1 2
给n个士兵的level值,问最少须要多少扫帚。
/*士兵要学骑扫帚。每一个士兵有一个level。 level高的能在同一把扫帚上教level低的怎么骑。一个人最多有一个老师。一个学生。也能够没有。 给n个士兵的level值,问最少须要多少扫帚。*/ #include <iostream> #include <cstdio> #include <algorithm> #include <cstring> using namespace std; struct node { node *next[10]; int Count; node() { for (int i=0; i<10; i++) next[i]=NULL; Count=0; } }; int Max; node *p,*root; void Insert(char *s) { p=root; for (int i=0; s[i]; i++) { int k=s[i]-'0'; if (p->next[k]==NULL) p->next[k]=new node(); p=p->next[k]; } p->Count++; if (p->Count>Max) Max=p->Count; } /*void Search(int *s) { p=root; for (int i=0;i<n;i++) { int k=s[i]-'0'; if (p->next[k]==NULL]) return ; p=p->next[k]; } }*/ int main() { int n,j; char ch[40]; while (~scanf("%d",&n)) { root=new node(); Max=0; for (int i=0; i<n; i++) { scanf("%s",ch); int len=strlen(ch); for (j=0; j<len; j++) { if (ch[j]!='0') break; } Insert(ch+j); } printf ("%d\n",Max); } return 0; }