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.
For each case, output the minimum number of broomsticks on a single line.
1
2
翻译题。。。
大意是给你n个数字，找出n个数字重复次数最多的次数。
用map超时，那么就用字典树，每一次输入数字就插入字典树中，然后用外部变量MAXH和次数比较，若MAXH小于该次数，MAXH=该次数，插入一遍后MAXH就为最终结果。
代码不解释：
1 #include <stdio.h>
2 #include <string.h>
3 #include <algorithm>
4 #include <map>
5 #include <iostream>
6 #include <string>
7 using namespace std;
8
9 int MAXH;
10 struct node{
11 int num;
12 struct node *next[10];
13 node(){
14 num=0;
15 memset(next,0,sizeof(next));
16 }
17 };
18
19 void insert(char *s,struct node &root)
20 {
21 struct node *p=&root;
22 int k=0;
23 while(s[k])
24 {
25 if(!p->next[s[k]-‘0‘]){
26 p->next[s[k]-‘0‘]=new node;
27 p=p->next[s[k]-‘0‘];
28 }
29 else p=p->next[s[k]-‘0‘];
30 k++;
31 }
32 p->num++;
33 if(MAXH<p->num)
34 MAXH=p->num;
35 }
36
37 main()
38 {
39 int n, i, j, k;
40 char a[50];
41 while(scanf("%d",&n)==1)
42 {
43 struct node root;
44
45 MAXH=-1;
46 for(i=0;i<n;i++)
47 {
48 scanf("%s",a);
49 k=0;
50 while(a[k]==‘0‘) k++;
51 if(k==strlen(a))
52 k--;
53 insert(a+k,root);
54 }
55
56 printf("%d\n",MAXH);
57 }
58 }
