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困难
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第 168 场周赛 Q4
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数组

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题目描述

给你 n 个盒子,每个盒子的格式为 [status, candies, keys, containedBoxes] ,其中:

  • 状态字 status[i]:整数,如果 box[i] 是开的,那么是 ,否则是
  • 糖果数 candies[i]: 整数,表示 box[i] 中糖果的数目。
  • 钥匙 keys[i]:数组,表示你打开 box[i] 后,可以得到一些盒子的钥匙,每个元素分别为该钥匙对应盒子的下标。
  • 内含的盒子 containedBoxes[i]:整数,表示放在 box[i] 里的盒子所对应的下标。

给你一个 initialBoxes 数组,表示你现在得到的盒子,你可以获得里面的糖果,也可以用盒子里的钥匙打开新的盒子,还可以继续探索从这个盒子里找到的其他盒子。

请你按照上述规则,返回可以获得糖果的 最大数目 

 

示例 1:

输入:status = [1,0,1,0], candies = [7,5,4,100], keys = [[],[],[1],[]], containedBoxes = [[1,2],[3],[],[]], initialBoxes = [0]
输出:16
解释:
一开始你有盒子 0 。你将获得它里面的 7 个糖果和盒子 1 和 2。
盒子 1 目前状态是关闭的,而且你还没有对应它的钥匙。所以你将会打开盒子 2 ,并得到里面的 4 个糖果和盒子 1 的钥匙。
在盒子 1 中,你会获得 5 个糖果和盒子 3 ,但是你没法获得盒子 3 的钥匙所以盒子 3 会保持关闭状态。
你总共可以获得的糖果数目 = 7 + 4 + 5 = 16 个。

示例 2:

输入: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]
输出:6
解释:
你一开始拥有盒子 0 。打开它你可以找到盒子 1,2,3,4,5 和它们对应的钥匙。
打开这些盒子,你将获得所有盒子的糖果,所以总糖果数为 6 个。

示例 3:

输入:status = [1,1,1], candies = [100,1,100], keys = [[],[0,2],[]], containedBoxes = [[],[],[]], initialBoxes = [1]
输出:1

示例 4:

输入:status = [1], candies = [100], keys = [[]], containedBoxes = [[]], initialBoxes = []
输出:0

示例 5:

输入:status = [1,1,1], candies = [2,3,2], keys = [[],[],[]], containedBoxes = [[],[],[]], initialBoxes = [2,1,0]
输出:7

 

提示:

  • 1 <= status.length <= 1000
  • status.length == candies.length == keys.length == containedBoxes.length == n
  • status[i] 要么是 0 要么是 1
  • 1 <= candies[i] <= 1000
  • 0 <= keys[i].length <= status.length
  • 0 <= keys[i][j] < status.length
  • keys[i] 中的值都是互不相同的。
  • 0 <= containedBoxes[i].length <= status.length
  • 0 <= containedBoxes[i][j] < status.length
  • containedBoxes[i] 中的值都是互不相同的。
  • 每个盒子最多被一个盒子包含。
  • 0 <= initialBoxes.length <= status.length
  • 0 <= initialBoxes[i] < status.length

解法

方法一:BFS

Python3

class Solution:
    def maxCandies(
        self,
        status: List[int],
        candies: List[int],
        keys: List[List[int]],
        containedBoxes: List[List[int]],
        initialBoxes: List[int],
    ) -> int:
        q = deque([i for i in initialBoxes if status[i] == 1])
        ans = sum(candies[i] for i in initialBoxes if status[i] == 1)
        has = set(initialBoxes)
        took = {i for i in initialBoxes if status[i] == 1}

        while q:
            i = q.popleft()
            for k in keys[i]:
                status[k] = 1
                if k in has and k not in took:
                    ans += candies[k]
                    took.add(k)
                    q.append(k)
            for j in containedBoxes[i]:
                has.add(j)
                if status[j] and j not in took:
                    ans += candies[j]
                    took.add(j)
                    q.append(j)
        return ans

Java

class Solution {
    public int maxCandies(
        int[] status, int[] candies, int[][] keys, int[][] containedBoxes, int[] initialBoxes) {
        int ans = 0;
        int n = status.length;
        boolean[] has = new boolean[n];
        boolean[] took = new boolean[n];
        Deque<Integer> q = new ArrayDeque<>();
        for (int i : initialBoxes) {
            has[i] = true;
            if (status[i] == 1) {
                ans += candies[i];
                took[i] = true;
                q.offer(i);
            }
        }
        while (!q.isEmpty()) {
            int i = q.poll();
            for (int k : keys[i]) {
                status[k] = 1;
                if (has[k] && !took[k]) {
                    ans += candies[k];
                    took[k] = true;
                    q.offer(k);
                }
            }
            for (int j : containedBoxes[i]) {
                has[j] = true;
                if (status[j] == 1 && !took[j]) {
                    ans += candies[j];
                    took[j] = true;
                    q.offer(j);
                }
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    int maxCandies(vector<int>& status, vector<int>& candies, vector<vector<int>>& keys, vector<vector<int>>& containedBoxes, vector<int>& initialBoxes) {
        int ans = 0;
        int n = status.size();
        vector<bool> has(n);
        vector<bool> took(n);
        queue<int> q;
        for (int& i : initialBoxes) {
            has[i] = true;
            if (status[i]) {
                ans += candies[i];
                took[i] = true;
                q.push(i);
            }
        }
        while (!q.empty()) {
            int i = q.front();
            q.pop();
            for (int k : keys[i]) {
                status[k] = 1;
                if (has[k] && !took[k]) {
                    ans += candies[k];
                    took[k] = true;
                    q.push(k);
                }
            }
            for (int j : containedBoxes[i]) {
                has[j] = true;
                if (status[j] && !took[j]) {
                    ans += candies[j];
                    took[j] = true;
                    q.push(j);
                }
            }
        }
        return ans;
    }
};

Go

func maxCandies(status []int, candies []int, keys [][]int, containedBoxes [][]int, initialBoxes []int) int {
	ans := 0
	n := len(status)
	has := make([]bool, n)
	took := make([]bool, n)
	var q []int
	for _, i := range initialBoxes {
		has[i] = true
		if status[i] == 1 {
			ans += candies[i]
			took[i] = true
			q = append(q, i)
		}
	}
	for len(q) > 0 {
		i := q[0]
		q = q[1:]
		for _, k := range keys[i] {
			status[k] = 1
			if has[k] && !took[k] {
				ans += candies[k]
				took[k] = true
				q = append(q, k)
			}
		}
		for _, j := range containedBoxes[i] {
			has[j] = true
			if status[j] == 1 && !took[j] {
				ans += candies[j]
				took[j] = true
				q = append(q, j)
			}
		}
	}
	return ans
}