Leetcode 857 -- 贪心 | 数学

jyyyx的算法博客2025-04-05 19:42

题目描述

雇佣 K 名工人的最低成本

思路

参考官方题解和这里。

代码1（正确）

复制代码

class Solution {
public:
    double mincostToHireWorkers(vector<int>& quality, vector<int>& wage, int k) {
        int n = wage.size();
        double res = 0, totalq = 0; // totalquality
        vector<int> h(n);
        for(int i = 0; i < n; i ++ )    h[i] = i;
        sort(h.begin(), h.end(), [&](int &a, int &b){
           return wage[a] * quality[b] < wage[b] * quality[a]; 
        });
        for(int i = 0; i < n; i ++ )    cout << h[i] << ' ';
        cout << endl;
        priority_queue<int, vector<int>, less<int>> q; // 降序
        for(int i = 0; i < k - 1; i ++ )
        {
            q.push(quality[h[i]]);
            totalq += quality[h[i]];
        }
        for(int i = k - 1; i < n; i ++ )
        {
            q.push(quality[h[i]]);
            totalq += quality[h[i]];
            double cur = totalq / (double)quality[h[i]] * wage[h[i]];
            if(res == 0 || cur < res)  res = cur;
            totalq -= q.top();
            q.pop();
        }
        
        return res;
    }
};

代码2（错误）

复制代码

class Solution {
public:
    typedef pair<double, int> PDI;
    double mincostToHireWorkers(vector<int>& quality, vector<int>& wage, int k) {
        int n = wage.size();
        double res = 0, totalq = 0; // totalquality
        vector<int> h(n);
        for(int i = 0; i < n; i ++ )    h[i] = i;
        sort(h.begin(), h.end(), [&](int &a, int &b){
           return quality[a] < quality[b];
        });
        
        priority_queue<PDI, vector<PDI>, less<PDI>> q; // 降序
        for(int i = 0; i < k - 1; i ++ )
        {
            int idx = h[i];
            totalq += quality[idx];
            q.push({(double)wage[i] / quality[i],idx});
        }
        for(int i = k - 1; i < n; i ++ )
        {
            int idx = h[i];
            q.push({(double)wage[idx] / quality[idx],idx});          
            totalq += quality[idx];
            auto t = q.top();
            cout << totalq << ' ';
            double cur = totalq * t.first;
            if(res == 0 || cur < res)   res = cur;
            totalq -= quality[t.second];
            q.pop();
        }
        
        return res;
    }
};