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Pretty classic greedy problem to work on. Here is how to approach it:
1. "the smallest team is as large as possible." actually means, team members should be distributed as average as possible
2. When you are to put member with skill-level ‘x‘, you find the team with least members (say n members), whose lowest skill-level is x + 1, and update the record.
With statement above, you can combine hashmap and min-heap to solve it:
#include <cmath> #include <cstdio> #include <vector> #include <map> #include <set> #include <string> #include <climits> #include <iostream> #include <algorithm> #include <functional> #include <queue> #include <unordered_map> #include <unordered_set> using namespace std; void go() { int n; cin >> n; vector<int> in(n); for (int i = 0; i < n; i++) cin >> in[i]; std::sort(in.begin(), in.end(), std::greater<int>()); // min-value: sizes in heap unordered_map<int, priority_queue<size_t, std::vector<size_t>, std::greater<size_t>>> rec; for (auto v : in) { if (rec.find(v + 1) == rec.end()) { rec[v].push(1); } else { auto it = rec.find(v + 1); size_t old_minsize = it->second.top(); it->second.pop(); if (it->second.size() == 0) { rec.erase(v + 1); } rec[v].push(old_minsize + 1); } } // Get min size size_t min_size = std::numeric_limits<size_t>::max(); for (auto &r : rec) min_size = std::min(min_size, r.second.top()); cout << (min_size == std::numeric_limits<size_t>::max() ? 0:min_size) << endl; } int main() { int t; cin >> t; while (t--) { go(); } return 0; }
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原文地址：http://www.cnblogs.com/tonix/p/4543483.html
