hdu1800 Flying to the Mars(字典树)
来源:互联网 发布:php设计表格 编辑:程序博客网 时间:2024/06/09 17:45
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1800
Flying to the Mars
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 14340 Accepted Submission(s): 4572
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);
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
410203004523434
Sample Output
12
题目大意: 士兵要学骑扫帚。每个士兵有一个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。也可以没有。给n个士兵的level值,问最少需要多少扫帚。
详见代码。
/*士兵要学骑扫帚。每个士兵有一个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。也可以没有。给n个士兵的level值,问最少需要多少扫帚。*/#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;}
1 0
- 【字典树】 hdu1800 Flying to the Mars
- hdu1800 Flying to the Mars (字典树)
- Hdu1800 - Flying to the Mars - 字典树
- HDU1800 Flying to the Mars【字典树】
- hdu1800 Flying to the Mars--字典树
- Flying to the Mars hdu1800 字典树
- hdu1800 Flying to the Mars(字典树)
- HDU1800 Flying to the Mars Tire树
- HDU1800 Flying to the Mars(贪心)
- HDU1800 Flying to the Mars
- HDU1800 Flying to the Mars
- HDU1800-Flying to the Mars
- hdu1800 Flying to the Mars
- Flying to the Mars(字典树)
- Flying to the Mars(hdu1800,水题排序)
- HDU1800——Flying to the Mars(map+贪心)
- hdu1800 Flying to the Mars(map水)
- hdu1800 Flying to the Mars(map)
- 黑马程序员_IO流之异常处理机制与File类
- iOS 8 AutoLayout与Size Class的基本用法
- hdoj 1533 Going Home 【最小费用最大流】【KM入门题】
- Util:ToolFor9Ge
- PAT 1062. Talent and Virtue (25)
- hdu1800 Flying to the Mars(字典树)
- JSON浅总
- 匿名项目连载(四)--->评论说说
- Nagios
- Table is 'read only'
- 一个Oracle循环例子
- 积跬步,聚小流------div模拟select,让select美美哒
- 用Quartz处理定时执行的任务
- jdk中path和classpath的区别