End up with More Teams UVA - 11088

End up with More Teams

Time Limit: 3000MS

Memory Limit: Unknown

64bit IO Format: %lld & %llu

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Description

 

I I U P C 2 0 06
Problem E: End up with More Teams
Input: standard input Output: standard output
The prestigious ICPC is here again. The coaches are busy selecting teams. Well this year, they have adopted a new policy. Contrary to traditional selection process, where few individual contests are held and the top three are placed in one team the next three in another and so on, this year the coaches decided to place members in such a way that the total number of promising teams is maximized. Promising teams are defined as a team having ability points of its members adding up to 20 or greater. Here ability point of a member denotes his capability as a programmer, the higher the better.
Input
There will be as many as 100 cases in the input file. Each case of input has two lines. The first line contains a positive integer, where n indicates the number contestants available for selection. The next line will contain n positive integers, each of which will be at most 30.  End of input is indicated by a value of 0 forn.
Output
For every case of input there will be one line of output. This line should contain the case number followed by the maximum number of promising teams that can be formed. Note that it is not mandatory to assign everyone in a team. Incase you don’t know, each team consists of exactly 3 members.
Constraints
-           n ≤ 15
Sample Input	Output for Sample Input
9 22 20 9 10 19 30 2 4 16 2 15 3 0	Case 1: 3 Case 2: 0

 #include <iostream>
 #include <stdio.h>
 #include <string.h>
 #include <algorithm>
 using namespace std;
 int a[],b[],n,m,maxa;
 bool cmp(int x,int y)
 {
     return x>y;
 }
 void dfs(int x,int cur)
 {
     int i,j,k,sum;
     if(maxa<cur)maxa=cur;
     if(maxa==m)
         return;
     for(i=x; i<n; i++)
     {
         if(!b[i])
             for(j=i+; j<n; j++)
             {
                 if(!b[j])
                 {
                     sum=a[i]+a[j];
                     for(k=j+; k<n; k++)
                     {
                         if(!b[k]&&sum+a[k]>=)break;
                     }
                     if(k<n)
                     {
                         b[i]=b[j]=b[k]=;
                         dfs(i+,cur+);
                         b[i]=b[j]=b[k]=;
                     }
                 }
             }
     }
 }
 int main()
 {
     int i,cas=;
     while(~scanf("%d",&n),n)
     {
         maxa=;
         memset(b,,sizeof(b));
         for(i=; i<n; i++)scanf("%d",&a[i]);
         sort(a,a+n);
         m=n/;
         dfs(n%,);
         printf("Case %d: %d\n",cas++,maxa);
     }
 }