HDU - 2018杭电ACM集训队单人排位赛 - 2 - Problem C. Team Match

Problem C Team Match

Time Limit: 2000/1500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 160    Accepted Submission(s): 55

Problem Description

The programming competition not only depends on the programmers, but also directed by the coaches. Mr Z is a coach who direct n players to take part in the Guangxi Province Collegiate Programming Contest. We assume that a team is consisted of 3 players whose ability is x, y, z respectively and x >= y >= z. Then the team’s total ability is 3 * x + 2 * y + 1 * z; And for a team, if its ability is not lower than the gold medal level m, the team will certainly win the gold medal. Mr Z would like to match teams to gain as many gold medals as possible, could you tell him how many gold medals it is?

 

Input

The first line is an integer T which indicates the case number.
And as for each case,  there will be 2 lines.
In the first line, there are 2 integers n m, which indicate the number of players, the gold medal level respectively. Please remember n is always the multiple of 3.
In the second line, there are n integers which represents everyone’s ability.
It is guaranteed that——
T is about 100.
for 100% cases, 1 <= n <= 15, 1 <= m <= 30, 1 <= a[i] <= 20.

 

Output

As for each case, you need to output a single line.
There should be an integer in the line which means the gold medal teams Mr Z could match.

 

Sample Input
2
6 18
3 3 3 4 2 2
6 7
1 1 1 1 1 1

 

Sample Output
2
0

解题思路：从小到大排序后，每次选择一个最小，一个最大，然后中间 a[i]（i 从小到大）循环寻找合适的中间数。在还没整个数组统计完时，如果当中有一次 return false，那么就肯定有一对已经没法执行，那么选择最小的尽量靠前的3个淘汰掉，再整个重来。

AC 代码

#include
#include

#define mem(a,b) memset(a,b,sizeof a);
#define INF 0x3f3f3f3f

using namespace std;

typedef long long ll;
int n,m,a[20];

bool solve(int x)
{
    int l=(n/3-x)*3+1, r=n, vis[20];
    mem(vis,0);
    while(l=m)
            {
                flag=1;
                vis[i]=1;
                break;
            }
        }

        if(!flag) return false;
        vis[l]=vis[r]=1;
        while(vis[l]) l++;
        while(vis[r]) r--;
    }
    return true;
}

int main()
{
    int T; scanf("%d",&T);
    while(T-- && ~scanf("%d%d",&n,&m))
    {
        for(int i=1;i<=n;i++) scanf("%d",&a[i]);
        sort(a+1,a+1+n);
        int ans=n/3;
        while(ans)
        {
            if(solve(ans)) break;
            else ans--;
        }
        printf("%d\n",ans);
    }

    return 0;
}