注意题目这句话，Once you have each type of candies in a box, you want to pack those boxes into larger boxes, until only one box remains.
两个box合并后必须放入更大一个盒子
题目的有点类似huffman的前部分，此题用堆去做，由于priority_queue是用堆实现的，故可以直接使用
每次从堆中选取最小的两个进行合并即可
#include <iostream> #include <queue> #include <vector> #include <algorithm> using namespace std; class BoxesDiv2{ public: int round_up(int x){ for(int i = 0; i <=10 ; ++ i){ if( 1<<i >= x ) return 1<<i; } return 1<<11; } int findSize(vector<int> candyCounts){ priority_queue<int,vector<int>,greater<int> > boxQueue; for(int i = 0 ; i < candyCounts.size(); ++ i){ boxQueue.push(round_up(candyCounts[i])); } while(boxQueue.size() != 1){ int a = boxQueue.top();boxQueue.pop(); int b = boxQueue.top();boxQueue.pop(); boxQueue.push(max(a,b)*2); } return boxQueue.top(); } };
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topcoder SRM 522 DIV2 BoxesDiv2
原文：http://www.cnblogs.com/xiongqiangcs/p/3766763.html