[2011山东省第二届ACM大学生程序设计竞赛]——The Android University ACM Team Selection Contest

The Android University ACM Team Selection Contest

Time Limit: 1000MS Memory limit: 65536K

题目: http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2162

题目描述

 Now it's 20000 A.D., and the androids also participate in the ACM Inter-national Collegiate Programming Contest (ACM/ICPC). In order to select the members of Android University ACM/ICPC Training Camp, a contest was held. There were N teams competing in the contest, among which there might be some teams whose members are all girls (they are called all-girls teams). Some of the N teams will be selected, then all the members of those teams are selected as the members of the training camp.

To be selected, one team has to solve at least one problem in the contest. The the top M teams who solved at least one problem are selected (If there are less than M teams solving at least one problem, they are all selected).

There is an bonus for the girls - if top M teams contains no all-girls teams,the highest ranked all-girls team is also selected (together with the M top teams), provided that they have solved at least one problem.

Recall that in an ACM/ICPC style contest, teams are ranked as following:

1. The more problems a team solves, the higher order it has.

2. If multiple teams have the same number of solved problems, a team with a smaller penalty value has a higher order than a team with a

larger penalty value.

Given the number of teams N, the number M defined above, and each team's name, number of solved problems, penalty value and whether it's an all-girls team, you are required to write a program to find out which teams are selected.

输入

 The input has multiple test cases. The first line of the input contains one integer C, which is the number of test cases.

Each test case begins with a line contains two integers, N (1 <= N <=10^4) and M (1 <= M <= N), separated by a single space. Next will be N lines, each of which gives the information about one specific competing team.Each of the N lines contains a string S (with length at most 30, and consists of upper and lower case alphabetic characters) followed by three integers, A(0 <= A <= 10), T (0 <= T <= 10) and P (0 <= P <= 5000), where S is the name of the team, A indicates whether the team is an all-girls team (it is not an all-girls team if Ai is 0, otherwise it is an all-girls team). T is the number of problems the team solved, and P is the penalty value of the team.

The input guarantees that no two teams who solved at least one problem have both the same T and P.

输出

 For each test case, print one line containing the case number (starting from 1). Then, output the selected teams' names by the order they appear in the input, one on each line. Print a blank line between the output for two test cases. Refer to the Sample Output section for details.

示例输入

3
5 3
AU001 0 0 0
AU002 1 1 200
AU003 1 1 30
AU004 0 5 500
AU005 0 7 1000
2 1
BOYS 0 10 1200
GIRLS 10 1 290
3 3
red 0 0 0
green 0 0 0
blue 0 1 30

示例输出

Case 1:
AU003
AU004
AU005
Case 2:
BOYS
GIRLS
Case 3:
blue
3
3

这道题是类似于模拟题吧,这套题目有两个模拟类型。

只要细心仔细,把每一个状态都想好,一遍AC不是梦。

题意就是模拟ACM比赛排名,

在一些队伍中选择几个队伍获奖。

这道题有几个注意点:

①只有做出题目的队伍,才能获奖。

②如果选出获奖的队伍中没有女生队伍,那么要将所有女队中排名第一的也要给奖。

③获奖排名,按照输入时的顺序输出队名。

④如果获奖队伍不够,先输出获奖的队伍,剩下的空缺用,奖项数量来代替输出。

例如最后一个样例:只有blue获奖,但是要选出3个队伍,所以剩下两个空缺,用3来代替输出。

⑤是否是女队,用bool型存储可以,但不能直接赋值给该类型,

比如: bool b;cin>>b;

输入10,则会出错。可以设置一个int型来输入,再强制转换类型。

例如: bool b;int t;  cin>>t;  b=bool(t);


其他的事项,在代码中有注释。

#include 
#include 
#include 
#include 
using namespace std;
struct Team
{
    string name;
    bool isgirl;
    int aproblem,atime,num;
}team[10010];
int n,m,total;

// 按照AC题目多少与时间排序
bool cmp_a(Team a,Team b)
{
    if(a.aproblem==b.aproblem)
    {
        if(a.atimeb.aproblem;
}
// 按序号大小排序
bool cmp_num(Team a,Team b)
{
    return a.num>test;
    for(k=1;k<=test;++k)
    {
//        memset(team,0,sizeof(team));
        // 输入
        cin>>n>>m;
        total=0;
        for(i=0;i>temp.name>>te>>temp.aproblem>>temp.atime;
            // 一道题没答出来,舍掉
            if(!temp.aproblem)   continue;
            team[total].name=temp.name;
            team[total].isgirl=bool(te);
            team[total].aproblem=temp.aproblem;
            team[total].atime=temp.atime;
            team[total].num=i+1;
            ++total;
        }
        // 排序
        sort(team,team+total,cmp_a);
        // 输出
        cout<<"Case "<


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