The Android University ACM Team Selection Contest

Time Limit: 1000ms   Memory limit: 65536K  有疑问？点这里^_^

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

35 3AU001 0 0 0AU002 1 1 200AU003 1 1 30AU004 0 5 500AU005 0 7 10002 1BOYS 0 10 1200GIRLS 10 1 2903 3red 0 0 0green 0 0 0blue 0 1 30

示例输出

Case 1:AU003AU004AU005Case 2:BOYSGIRLSCase 3:blue33

提示

 

来源

山东省第二届ACM大学生程序设计竞赛

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　　模拟题。

　　代码：

1 #include 
      
        2 #include 
       
         3 using namespace std; 4 struct Team{ 5     char s[31]; 6     int a,b,c; 7 }team[10001]; 8 bool sel[10001]; 9 void solve(int n,int m,int cnt) //共n个队，输出m个队10 {11     int i,j,num=999999999;12     bool f = false; //是否有女队13     for(i=1;i<=m;i++){14         int Max=0,t;15         for(j=1;j<=n;j++)   //找到最大的16             if(team[j].b>Max && team[j].b<=num && !sel[j]){17                 Max = team[j].b;18                 t = j;19             }20         num = Max;21         if(num<1) break;22         //找到与这个值相等的最小的值23         for(j=1;j<=n;j++){24             if(team[j].b==num && team[j].c
        
         Max && !sel[j]){    //女队35                 Max = team[i].b;36                 t = i;37             }38         //寻找做出这个题数的罚时最少的女队39         if(Max!=0){40             for(i=t;i<=n;i++){41                 if(team[i].a!=0 && team[i].b==team[t].b && team[i].c
         
          >N;55     for(i=1;i<=N;i++){56         int n,m;57         memset(sel,0,sizeof(sel));58         cin>>n>>m;59         for(j=1;j<=n;j++)   //input60             cin>>team[j].s>>team[j].a>>team[j].b>>team[j].c;61         //sort(team+1,team+n+1,cmp);62         cout<<"Case "<
          <<':'<
         
        
       
      

 

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