LeetCode #1066. Campus Bikes II

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在一个二维网格的校园中，存在N名工人和M辆自行车，N小于等于M。每名工人和每辆自行车都是网格上的一个坐标点。本算法通过深度优先搜索和回溯策略，为每个工人分配一辆独特的自行车，目标是最小化所有工人与其分配自行车之间的曼哈顿距离总和。

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题目描述：

On a campus represented as a 2D grid, there are N workers and M bikes, with N <= M. Each worker and bike is a 2D coordinate on this grid.

We assign one unique bike to each worker so that the sum of the Manhattan distances between each worker and their assigned bike is minimized.

The Manhattan distance between two points p1 and p2 is Manhattan(p1, p2) = |p1.x - p2.x| + |p1.y - p2.y|.

Return the minimum possible sum of Manhattan distances between each worker and their assigned bike.

Example 1:

Input: workers = [[0,0],[2,1]], bikes = [[1,2],[3,3]]
Output: 6
Explanation: 
We assign bike 0 to worker 0, bike 1 to worker 1. The Manhattan distance of both assignments is 3, so the output is 6.

Example 2:

Input: workers = [[0,0],[1,1],[2,0]], bikes = [[1,0],[2,2],[2,1]]
Output: 4
Explanation: 
We first assign bike 0 to worker 0, then assign bike 1 to worker 1 or worker 2, bike 2 to worker 2 or worker 1. Both assignments lead to sum of the Manhattan distances as 4.

Note:

0 <= workers[i][0], workers[i][1], bikes[i][0], bikes[i][1] < 1000
All worker and bike locations are distinct.
1 <= workers.length <= bikes.length <= 10

class Solution {
public:
    int assignBikes(vector<vector<int>>& workers, vector<vector<int>>& bikes) {
        dist=INT_MAX;
        int cur=0;
        DFS(workers,bikes,0,cur,0);
        return dist;
    }
    
    void DFS(vector<vector<int>>& workers, vector<vector<int>>& bikes, int mask, int cur, int i)
    {
        if(cur>=dist) return;
        if(i==workers.size())
        {
            dist=min(dist,cur);
            return;
        }

        for(int j=0;j<bikes.size();j++)
        {
            if(((1<<j)&mask)>0) continue;
            mask+=(1<<j);
            DFS(workers,bikes,mask,cur+manhattan_dist(workers[i],bikes[j]),i+1);
            mask-=(1<<j);
        }
    }
    
    int manhattan_dist(const vector<int>& a, const vector<int>& b)
    {
        return abs(a[0]-b[0])+abs(a[1]-b[1]);
    }
    
private:
    int dist;
};