comments | difficulty | edit_url | rating | source | tags | |||
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true |
困难 |
1824 |
第 168 场周赛 Q4 |
|
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 1 ,否则是 0 。 - 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
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
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;
}
}
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;
}
};
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
}