Hackerrank--Team Formation

最新推荐文章于 2022-07-31 12:43:30 发布

转载最新推荐文章于 2022-07-31 12:43:30 发布 · 120 阅读

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原文链接：http://www.cnblogs.com/Stomach-ache/p/3947530.html

本文介绍了一个关于团队组建的问题，目标是在确保团队内成员技能连续且无重复的前提下，最大化最小团队规模。通过一种贪心策略，文章提供了一种有效解决方法，并附带样例输入输出及代码实现。

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题目链接

For an upcoming programming contest, Roy is forming some teams from the

Roy knows the skill level of each contestant. To make the teams work as a unit, he should ensure that there is no skill gap between the contestants of the same team. In other words, if the skill level of a contestant is

It is clear that more the number of contestants in a team, more the problems they can attempt at a time. So Roy wants to form teams such that the size of the smallest possible team is maximized.

Input Format

The first line of input contains

Each case contains an integer

The total number of contestants in all cases will not exceed

Output Format

For each test case, print the output in a separate line.

Sample Input
4  
7 4 5 2 3 -4 -3 -5  
1 -4  
4 3 2 3 1  
7 1 -2 -3 -4 2 0 -1  
Sample Output
3
1
1
7
Explanation

For the first case, Roy can form two teams: one with contestants with skill levels{-4,-3,-5} and the other one with {4,5,2,3}. The first group containing 3 members is the smallest.

In the second case, the only team is {-4}

In the third case, the teams are {3} , {1,2,3}, the size of the smaller group being 1.

In the last case, you can build a group containing all the contestants. The size of the group equals the total number of contestants.

思路：贪心。尽量减少分组的个数。

如果存在以当前这个数-1为结尾的组，那么就找到个数最少的一个将当前这个数加在后面，否则增加一个组，以当前数为结尾。

Accepted Code:

 1 using namespace std;
 2 #include <bits/stdc++.h>
 3 
 4 #define sgn(x,y) ((x)+eps<(y)?-1:((x)>eps+(y)?1:0))
 5 #define rep(i,n) for(auto i=0; i<(n); i++)
 6 #define mem(x,val) memset((x),(val),sizeof(x));
 7 #define rite(x) freopen(x,"w",stdout);
 8 #define read(x) freopen(x,"r",stdin);
 9 #define all(x) x.begin(),x.end()
10 #define sz(x) ((int)x.size())
11 #define sqr(x) ((x)*(x))
12 #define pb push_back
13 #define clr clear()
14 #define inf (1<<28)
15 #define ins insert
16 #define xx first
17 #define yy second
18 #define eps 1e-9
19 
20 int main(void) {
21     ios_base::sync_with_stdio(0);
22     int test;
23     cin >> test;
24     while ( test-- ) {
25         map<int, priority_queue<int, vector<int>, greater<int> > > val;
26         int n;
27         cin >> n;
28         vector<int> vec(n);
29         rep(i, n) {
30             cin >> vec[i];
31         }
32         sort( vec.begin(), vec.end() );
33         
34         rep(i, n) {
35             int tmp = vec[i];
36             int now = 0;
37             auto it = val.find(tmp - 1);
38             if (it != val.end() && it->yy.size()) {
39                 now = it->yy.top(); 
40                 it->yy.pop();
41             } 
42             now++;
43             val[tmp].push(now);
44         }
45         int ans = INT_MAX;
46         for ( auto x : val ) if ( x.second.size() )
47             ans = min( ans, x.second.top() );
48         if (ans == INT_MAX) ans = 0;
49         cout << ans << endl;
50     }
51     
52     return 0;
53 }

转载于:https://www.cnblogs.com/Stomach-ache/p/3947530.html