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.
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.
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
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 <iostream>
#include <algorithm>
#include <string>
#include <string.h>
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.atime<b.atime)
return 1;
else return 0;
}
else
return a.aproblem>b.aproblem;
}
// 按序号大小排序
bool cmp_num(Team a,Team b)
{
return a.num<b.num;
}
// 判断前m个队伍中是否有女队
bool have_girl(void)
{
int i;
for(i=0;i<m;++i)
if(team[i].isgirl)
return true;
return false;
}
int main()
{
Team temp;
int i,k,test,te;
cin>>test;
for(k=1;k<=test;++k)
{
// memset(team,0,sizeof(team));
// 输入
cin>>n>>m;
total=0;
for(i=0;i<n;++i)
{
cin>>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 "<<k<<":"<<endl;
// 如果做出题目的队伍数量不足M个
if(total<=m)
{
sort(team,team+total,cmp_num);
for(i=0;i<total;++i)
cout<<team[i].name<<endl;
for(;i<m;++i)
cout<<m<<endl;
cout<<endl;
continue;
}
// 如果前m个队伍中没有女队
if( !have_girl() )
{
for(i=m;i<total;++i)
{
if(team[i].isgirl)
{
team[m].name=team[i].name;
team[m].num=team[i].num;
break;
}
}
sort(team,team+m+1,cmp_num);
for(i=0;i<=m;++i)
cout<<team[i].name<<endl;
cout<<endl;
}
else
{
sort(team,team+m,cmp_num);
for(i=0;i<m;++i)
cout<<team[i].name<<endl;
cout<<endl;
}
}
return 0;
}[2011山东省第二届ACM大学生程序设计竞赛]——The Android University ACM Team Selection Contest,布布扣,bubuko.com
[2011山东省第二届ACM大学生程序设计竞赛]——The Android University ACM Team Selection Contest
原文：http://blog.csdn.net/lttree/article/details/24180281