LeetCode-1298. Maximum Candies You Can Get from Boxes

LeetCode-1298. Maximum Candies You Can Get from Boxes

Given ​​n​​​ boxes, each box is given in the format ​​[status, candies, keys, containedBoxes]​​ where:

​​status[i]​​: an integer which is1if​​box[i]​​ is open and0if​​box[i]​​ is closed.​​candies[i]​​​: an integer representing the number of candies in​​box[i]​​.​​keys[i]​​​: an array contains the indices of the boxes you can open with the key in​​box[i]​​.​​containedBoxes[i]​​​: an array contains the indices of the boxes found in​​box[i]​​.

You will start with some boxes given in ​​initialBoxes​​ array. You can take all the candies in any open box and you can use the keys in it to open new boxes and you also can use the boxes you find in it.

Return the maximum number of candies you can get following the rules above.

Example 1:

Input: status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]Output: 16Explanation: You will be initially given box 0. You will find 7 candies in it and boxes 1 and 2. Box 1 is closed and you don't have a key for it so you will open box 2. You will find 4 candies and a key to box 1 in box 2.In box 1, you will find 5 candies and box 3 but you will not find a key to box 3 so box 3 will remain closed.Total number of candies collected = 7 + 4 + 5 = 16 candy.

Example 2:

Input: status = [1,0,0,0,0,0], candies = [1,1,1,1,1,1], keys = [[1,2,3,4,5],[],[],[],[],[]], containedBoxes = [[1,2,3,4,5],[],[],[],[],[]], initialBoxes = [0]Output: 6Explanation: You have initially box 0. Opening it you can find boxes 1,2,3,4 and 5 and their keys. The total number of candies will be 6.

Example 3:

Input: status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]Output: 1

Example 4:

Input: status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []Output: 0

Example 5:

Input: status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]Output: 7

Constraints:

​​1 <= status.length <= 1000​​​​status.length == candies.length == keys.length == containedBoxes.length == n​​​​status[i]​​​ is​​0​​​ or​​1​​.​​1 <= candies[i] <= 1000​​​​0 <= keys[i].length <= status.length​​​​0 <= keys[i][j] < status.length​​All values in​​keys[i]​​ are unique.​​0 <= containedBoxes[i].length <= status.length​​​​0 <= containedBoxes[i][j] < status.length​​All values in​​containedBoxes[i]​​ are unique.Each box is contained in one box at most.​​0 <= initialBoxes.length <= status.length​​​​0 <= initialBoxes[i] < status.length​​

​​题解：​​

​​bfs，使用两个数组保存已有箱子和钥匙，每次开完箱子都进行查找是否有钥匙跟箱子匹配。​​

class Solution {public: int maxCandies(vector& status, vector& candies, vector>& keys, vector>& containedBoxes, vector& initialBoxes) { int n = status.size(); queue q; vector box(n, false); vector key(n, false); int res = 0; for (int i = 0; i < initialBoxes.size(); i++) { if (status[initialBoxes[i]] == 1) { q.push(initialBoxes[i]); } if (status[initialBoxes[i]] == 0) { box[initialBoxes[i]] = true; } } while (q.empty() == false) { int t = q.front(); q.pop(); res += candies[t]; for (int i = 0; i < keys[t].size(); i++) { key[keys[t][i]] = true; } for (int i = 0; i < containedBoxes[t].size(); i++) { if (status[containedBoxes[t][i]] == 1) { q.push(containedBoxes[t][i]); } if (status[containedBoxes[t][i]] == 0) { box[containedBoxes[t][i]] = true; } } for (int i = 0; i < n; i++) { if (key[i] == true && box[i] == true) { q.push(i); key[i] = false; box[i] = false; } } } return res; }};

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