PAT1125:Chain the Ropes

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原文链接：http://www.cnblogs.com/0kk470/p/7642984.html

本文介绍了一种通过不断合并最短绳段来形成最长可能绳子的算法。采用类似哈夫曼树的贪心策略，每次选取长度最短的两段绳子进行合并，并将合并后的绳子作为新的一段参与后续合并过程。

1125. Chain the Ropes (25)

时间限制

200 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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.

Sample Input:

8
10 15 12 3 4 13 1 15

Sample Output:

14

思路

可以用哈夫曼树的思路（其实本质就是一个贪婪算法）：优先选两个最小的绳子合成新的绳子，然后循环即可。
所以为了得到最长的绳子，那么只需要对输入数据升序排下序即可。

代码

#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;
  }
}