Flying to the Mars(Hash)
来源:互联网 发布:阿里云邮箱手机客户端 编辑:程序博客网 时间:2024/05/21 22:44
The Suspects
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);
Output
For each case, output the minimum number of broomsticks on a single line.
Sample Input
410203004523434
Sample Output
12
Code:
#include <iostream>#include <cstdio>#include <cstring>#define mem(a,b) memset(a,b,sizeof(a))const int maxn=7003;int hash[maxn],count[maxn];int maxit,n;inline int ELFhash(char *key){ unsigned long h=0; unsigned long g; while(*key) { h=(h<<4)+*key++; g=h*0xf0000000L; if(g)h^=g>>24; h&=~g; } return h;}inline void hashit(char *str){ int k,t; while(*str=='0') str++; k=ELFhash(str); t=k%maxn; while(hash[t]!=k&&hash[t]!=-1) t=(t+10)%maxn; if(hash[t]==-1) count[t]=1,hash[t]=k; else if(++count[t]>maxit) maxit=count[t];}int main(){ char str[100]; while(scanf("%d",&n)!=EOF) { mem(hash,-1); for(maxit=1,gets(str);n>0;n--) { gets(str); hashit(str); } printf("%d\n",maxit); } return 0;}
阅读全文
0 0
- Flying to the Mars(Hash)
- hdu 1800 flying to the Mars(BKDR Hash)
- HDU 1800 Flying to the Mars(字符串Hash)
- Flying to the Mars (hash) 1800 hdu
- HDOJ 1800 Flying to the Mars hash
- Flying to the Mars
- Flying to the Mars
- Flying to the Mars
- Flying to the Mars
- Flying to the Mars
- Flying to the Mars
- Flying to the Mars
- Flying to the Mars
- Flying to the Mars
- Flying to the Mars
- HDU 1800 Flying to the Mars Trie或者hash
- 【邻接表字符串Hash】【HDU1800】Flying to the Mars
- HDU 1800 Flying to the Mars( map or hash )
- POJ2480 Longge's problem 欧拉函数应用
- vue axios全局配置问题
- ANR大全
- 使用主题预加载背景
- LeetCode 118 Pascal's Triangle
- Flying to the Mars(Hash)
- 测试中,一些不常见的问题汇总
- 计算机类|期刊】SCI期刊专刊截稿信息4条
- selenium 环境部署
- 全局唯一ID生成器浅析IdGen (1)
- 字符转化为数字
- (5)关于Flume内存溢出的问题,此时会报各种莫名奇妙的异常,异常如下 [html] view plain copy print? <span style="font-size:18px;">org
- 移动端兼容性问题解决方案
- 【DButils学习之】利用ResultSetHandler各实现类来处理查询结果