标签：leetcode worker order back 压缩 span style 剪枝 个数
Backtracing + Pruning (TLE)
最容易的解法是 backtracing，但是即使加了剪枝，还是会超时。
class Solution { public: int min_dist=INT_MAX; int assignBikes(vector<vector<int>>& workers, vector<vector<int>>& bikes) { vector<bool> bikeUsed(bikes.size(),false); dfs(workers,bikes,0,bikeUsed,0); return min_dist; } // assign a bike to employee i void dfs(vector<vector<int>>& workers, vector<vector<int>>& bikes, int i, vector<bool> &bikeUsed, int sum){ if (i==workers.size()){ min_dist = min(min_dist, sum); return; } if (sum>min_dist) return; // pruning for (int j=0;j<bikes.size();++j){ if (bikeUsed[j]) continue; bikeUsed[j] = true; dfs(workers,bikes,i+1,bikeUsed,sum+dist(workers[i],bikes[j])); bikeUsed[j] = false; } } int dist(vector<int> worker, vector<int> bike){ return abs(worker[0]-bike[0])+abs(worker[1]-bike[1]); } };
Backtracing + Memoization
加一个memo减少递归次数。由于上面的写法的递归函数返回void，要修改为有返回值的写法才能运用memo。
让 dfs 返回值为最小的距离和，然后将 bikeUsed 作为判断标识。由于bike的数量<=10，我们可以用int二进制对这个vector进行压缩作为key。这里只需要 bikeUsed 的原因是，从这个数组我们能得知分配了多少车，而人又是根据序号依次分配的，所以能得出哪些人还没有车。所以 bikeUsed 数组就能唯一确定一个子问题的状态。
class Solution { public: int assignBikes(vector<vector<int>>& workers, vector<vector<int>>& bikes) { vector<bool> bikeUsed(bikes.size(),false); unordered_map<int,int> memo; return dfs(workers,bikes,0,bikeUsed,memo); } // assign a bike to employee i int dfs(vector<vector<int>>& workers, vector<vector<int>>& bikes, int i, vector<bool> &bikeUsed, unordered_map<int,int> &memo){ if (i==workers.size()) return 0; int compressed_bikeUsed=compress(bikeUsed); if (memo.count(compressed_bikeUsed)) return memo[compressed_bikeUsed]; int min_dist=INT_MAX; for (int j=0;j<bikes.size();++j){ if (bikeUsed[j]) continue; bikeUsed[j] = true; min_dist = min(min_dist, dfs(workers,bikes,i+1,bikeUsed,memo) + dist(workers[i],bikes[j])); bikeUsed[j] = false; } return memo[compressed_bikeUsed]=min_dist; } int dist(vector<int> worker, vector<int> bike){ return abs(worker[0]-bike[0])+abs(worker[1]-bike[1]); } int compress(vector<bool> bikeUsed){ int res=0; for (int i=0;i<bikeUsed.size();++i){ res <<= 1; res |= bikeUsed[i]; } return res; } };
DP
既然可以 Backtracing + Memoization，那么也肯定可以 dp 来做。面试中遇到还是 backtracing 不容易出错。
待补...
Reference
https://leetcode.com/problems/campus-bikes-ii/discuss/305218/DFS-+-Pruning-And-DP-Solution
LeetCode 1066. Campus Bikes II
标签：leetcode worker order back 压缩 span style 剪枝 个数
原文地址：https://www.cnblogs.com/hankunyan/p/11261779.html
