LeetCode 1057. Campus Bikes

直接根据最短distance greedy做。

用了优先队列，C++ less<type> 是大的先，所以我们可以重载<符号，把逻辑反过来。

class Solution {
public:
    struct dist_pair{
        int dist;
        int worker;
        int bike;
        bool operator<(const dist_pair &a) const {
            if (dist!=a.dist) return dist>a.dist;
            else if (worker!=a.worker) return worker>a.worker;
            return bike>a.bike;
        }
    };
    
    vector<int> assignBikes(vector<vector<int>>& workers, vector<vector<int>>& bikes) {
        priority_queue<dist_pair> q;
        for (int i=0;i<workers.size();++i){
            for (int j=0;j<bikes.size();++j){
                int d = abs(workers[i][0] - bikes[j][0]) + abs(workers[i][1] - bikes[j][1]);
                q.push({d,i,j});    
            }
        }
        
        vector<int> assign(workers.size(),-1);
        vector<bool> visitedBike(bikes.size(),false);
        while (!q.empty()){
            auto [d,i,j] = q.top(); q.pop();
            if(assign[i]==-1 && visitedBike[j]==false){
                visitedBike[j] = true;
                assign[i] = j;
            }   
        }
        return assign;
    }
};

第二种写法不用结构体，节省了空间复杂度，但是每次cmp的时候都要计算dist，但是时间会翻倍。

这里采用了 lambda funtion，注意 priority_queue 里 lambda 的写法。

class Solution {
public:
    vector<int> assignBikes(vector<vector<int>>& workers, vector<vector<int>>& bikes) {
        auto cmp=[&](const pair<int,int> &a, const pair<int,int> &b){
            int dist_a=dist(a.first,a.second,workers,bikes);
            int dist_b=dist(b.first,b.second,workers,bikes);
            if (dist_a!=dist_b) return dist_a>dist_b;
            if (a.first!=b.first) return a.first>b.first;
            return a.second>b.second;
        };
        priority_queue<pair<int,int>,vector<pair<int,int>>,decltype(cmp)> q(cmp);
        
        for (int i=0;i<workers.size();++i)
            for (int j=0;j<bikes.size();++j)
                q.push({i,j});
        
        vector<int> assign(workers.size(),-1);
        vector<bool> visitedBike(bikes.size(),false);
        while (!q.empty()){
            auto [i,j] = q.top(); q.pop();
            if(assign[i]==-1 && visitedBike[j]==false){
                visitedBike[j] = true;
                assign[i] = j;
            }
        }
        return assign;
    }
    
    int dist(int i, int j, vector<vector<int>>& workers, vector<vector<int>>& bikes){
        return abs(workers[i][0]-bikes[j][0])+abs(workers[i][1]-bikes[j][1]);
    }
};

时间复杂度：mnlog(mn)，因为有mn对pair。