1125. Chain the Ropes (25)

Given some segments of rope, you are supposed to chain them into one rope. Each time you may only fold two segments into loops and chain them into one piece, as shown by the figure. The resulting chain will be treated as another segment of rope and can be folded again. After each chaining, the lengths of the original two segments will be halved.

Your job is to make the longest possible rope out of N given segments.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (2 <= N <= 10⁴). Then N positive integer lengths of the segments are given in the next line, separated by spaces. All the integers are no more than 10⁴

Output Specification:

For each case, print in a line the length of the longest possible rope that can be made by the given segments. The result must be rounded to the nearest integer that is no greater than the maximum length.

8
10 15 12 3 4 13 1 15

14

思路

可以用哈夫曼樹的思路（其實本質就是一個貪婪算法）：優先選兩個最小的繩子合成新的繩子，然後循環即可。
所以為了得到最長的繩子，那麽只需要對輸入數據升序排下序即可。

代碼
 

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
int main()
{
  int N;
  while(cin >> N)
  {
   vector<int> ropes(N);
   for(int i = 0;i < N;i++)
   {
      cin >> ropes[i];
   }
   sort(ropes.begin(),ropes.end());
   int length = 0;
   for(int i = 0;i < N ;i++)
   {
      if(i == 0)
      {
        length = ropes[i];
        continue;
      }
      length = (length + ropes[i])/2;
   }
   cout << length << endl;
  }
}

PAT1125:Chain the Ropes