zoukankan      html  css  js  c++  java
  • HDU 1800 Flying to the Mars

    Flying to the Mars
    Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 18245 Accepted Submission(s): 5881


    Problem Description ![PPF ARMY](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.


    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

    4
    10
    20
    30
    04
    5
    2
    3
    4
    3
    4


    Sample Output

    1
    2


    Author PPF@JLU
    解析:绕了一大圈,其实是求相同级别的人最多是几个(注意去首0)。
    ``` #include #include #include using namespace std;

    map<string, int> m;
    map<string, int>::iterator it;

    int main()
    {
    int n;
    char s[35];
    while(~scanf("%d", &n)){
    while(n--){
    scanf("%s", s);
    int cnt = 0;
    for(int i = 0; s[i] == '0'; ++i)
    ++cnt;
    if(cnt != 0){
    for(int i = 0; s[i] != ''; ++i)
    s[i] = s[i+cnt];
    }
    ++m[s];
    }
    int res = 0;
    for(it = m.begin(); it != m.end(); ++it){
    if(it->second > res)
    res = it->second;
    }
    printf("%d ", res);
    m.clear();
    }
    return 0;
    }

  • 相关阅读:
    HTML5侧滑聊天面板
    HTML5世界地图
    BZOJ_1042_[HAOI2008]硬币购物_容斥原理+背包
    BZOJ_1342_[Baltic2007]Sound静音问题_单调队列
    BZOJ_2343_[Usaco2011 Open]修剪草坪 _单调队列_DP
    BZOJ_2595_[Wc2008]游览计划_斯坦纳树
    BZOJ_5180_[Baltic2016]Cities_ 斯坦纳树
    BZOJ_4006_[JLOI2015]管道连接_斯坦纳树
    51nod_1412_AVL树的种类_动态规划
    BZOJ_3143_[Hnoi2013]游走_期望DP+高斯消元
  • 原文地址:https://www.cnblogs.com/inmoonlight/p/6034079.html
Copyright © 2011-2022 走看看