hdu1800fly to Mars字典树
来源:互联网 发布:美国下饺子知乎 编辑:程序博客网 时间:2024/06/01 16:08
其实可以用贪心,要不是数据太大==可是字典树更费空间==另:结构体可以有构造函数
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 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
#include <iostream>#include<cstdio>#include <iostream>#include<cstring>#include<cstdio>#include<cstdlib>using namespace std;int m,n;struct node{ node *next[10]; int num; node() { for(int i=0;i<10;i++) next[i]=0; num=0; }}*root;void inst(char s[]){ int k=0; node *p=root; while(s[k]!='\0') { if(!p->next[s[k]-'0']) p->next[s[k]-'0']=new node(); p=p->next[s[k]-'0']; k++; } p->num++; if(p->num>m) m=p->num;}int main(){ while(cin>>n) { m=0; root=new node(); for(int i=0;i<n;i++) { char s[33]; cin>>s; int j=0; while(s[j]=='0') j++; inst(s+j); } cout<<m<<'\n'; } return 0;}
0 0
- hdu1800fly to Mars字典树
- 【字典树】 hdu1800 Flying to the Mars
- hdu1800 Flying to the Mars (字典树)
- HDU1800Flying to the Mars(字典树)
- hdu1800Flying 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(字典树)
- hdu1800 Flying to the Mars--字典树
- Flying to the Mars hdu1800 字典树
- 【练习04】 字典树 1002 Flying to the Mars
- hdu 1800 Flying to the Mars(字典树)
- [字典树、map] HDU 1800 - Flying to the Mars
- hdu 1800 Flying to the Mars 字典树
- HDU 1800 Flying to the Mars(字典树)
- hdu1800 Flying to the Mars(字典树)
- hdu 1800 Flying to the Mars 字典树
- hdu 1800 Flying to the Mars 字典树
- Android屏幕适配全攻略(最权威的官方适配指导)
- SharePreferences的用法
- Core Data的笔记1
- hdu4002 Find the maximum 欧拉函数
- utils util
- hdu1800fly to Mars字典树
- github图片加载框架glide使用介绍
- 【牛腩新闻发布系统】系统设计说明书
- JAX-RS基础 for RESTful Web Service
- javascript 基础
- Android ListView 小笔记
- <Linux>文本编辑
- C++作业4.18
- 作业一 目标二 从命令行输入一个参数(指定目录或文件),输出该目录下指定类型文件(.cs, .java)的个