3112. 访问消失节点的最少时间
给你一个二维数组 edges 表示一个 n 个点的无向图，其中 edges[i] = [ui, vi, lengthi] 表示节点 ui 和节点 vi 之间有一条需要 lengthi 单位时间通过的无向边。
同时给你一个数组 disappear ，其中 disappear[i] 表示节点 i 从图中消失的时间点，在那一刻及以后，你无法再访问这个节点。
注意，图有可能一开始是不连通的，两个节点之间也可能有多条边。
请你返回数组 answer ，answer[i] 表示从节点 0 到节点 i 需要的 最少 单位时间。如果从节点 0 出发 无法 到达节点 i ，那么 answer[i] 为 -1 。
1 <= n <= 5 * 10**4
0 <= edges.length <= 10**5
edges[i] == [ui, vi, lengthi]
0 <= ui, vi <= n - 1
1 <= lengthi <= 10**5
disappear.length == n
1 <= disappear[i] <= 10**5
这题一看，对我来说是不会的一题，然而我看了灵神的解后，发现Python的解会用到堆的操作，Ruby是不带优先队列原生API支持的，
但Python封装heap的细节没暴露出来，然后我看到了Java解里对优先队列的排序有明确的定义，我可以尝试用Java解的思路让chatgpt
3.5来写。废话不多说，我开始用蹩脚英文来求助chatgpt3.5了。
I ：rewrite code in ruby
chatgpt:Sure, please provide the code you'd like to have rewritten in Ruby.
I :
class Solution {
public int[] minimumTime(int n, int[][] edges, int[] disappear) {
List<int[]>[] g = new ArrayList[n]; // 稀疏图用邻接表
Arrays.setAll(g, i -> new ArrayList<>());
for (int[] e : edges) {
int x = e[0];
int y = e[1];
int wt = e[2];
g[x].add(new int[]{y, wt});
g[y].add(new int[]{x, wt});
}
int[] dis = new int[n];
Arrays.fill(dis, -1);
dis[0] = 0;
PriorityQueue<int[]> pq = new PriorityQueue<>((a, b) -> (a[0] - b[0]));
pq.offer(new int[]{0, 0});
while (!pq.isEmpty()) {
int[] p = pq.poll();
int dx = p[0];
int x = p[1];
if (dx > dis[x]) { // x 之前出堆过
continue;
}
for (int[] e : g[x]) {
int y = e[0];
int newDis = dx + e[1];
if (newDis < disappear[y] && (dis[y] < 0 || newDis < dis[y])) {
dis[y] = newDis; // 更新 x 的邻居的最短路
pq.offer(new int[]{newDis, y});
}
}
}
return dis;
}
}
chatgpt:
class Solution
def minimum_time(n, edges, disappear)
g = Array.new(n) { [] }
edges.each do |e|
x, y, wt = e
g[x] << [y, wt]
g[y] << [x, wt]
end
dis = Array.new(n, -1)
dis[0] = 0
pq = [[0, 0]]
until pq.empty?
dx, x = pq.shift
next if dx > dis[x]
g[x].each do |e|
y, wt = e
new_dis = dx + wt
if new_dis < disappear[y] && (dis[y] < 0 || new_dis < dis[y])
dis[y] = new_dis
pq << [new_dis, y]
end
end
pq.sort_by!(&:first)
end
dis
end
end
Explanation:
In Ruby, the Array.new(n) { [] } creates an array of size n with each element being an empty array (similar to Java's new ArrayList[n] and Arrays.setAll combined).
The each method is used to iterate over the edges array.
The Array.new(n, -1) creates an array of size n initialized with -1.
The priority queue is implemented using an array (pq) and sorted by the first element to maintain the same behavior as the Java PriorityQueue.
The sort_by!(&:first) sorts the array in-place based on the first element of each sub-array to mimic the priority queue behavior in the original Java code.
I:please implement a high performance heap using in the code
chatgpt:
Sure, I'll implement a high-performance heap in Ruby for the priority queue operations. Ruby's standard library includes the Heap class within the algorithms gem which provides an efficient implementation of heaps. However, since the algorithms gem is not always available, I will implement a simple binary heap directly in the code.
Here's the updated code with a custom binary heap implementation:
class MinHeap
def initialize
@heap = []
end
def push(val)
@heap << val
sift_up(@heap.size - 1)
end
def pop
return if @heap.empty?
swap(0, @heap.size - 1)
min = @heap.pop
sift_down(0)
min
end
def empty?
@heap.empty?
end
private
def sift_up(index)
parent = (index - 1) / 2
return if index <= 0 || @heap[parent] <= @heap[index]
swap(index, parent)
sift_up(parent)
end
def sift_down(index)
left_child = 2 * index + 1
right_child = 2 * index + 2
min_index = index
if left_child < @heap.size && @heap[left_child] < @heap[min_index]
min_index = left_child
end
if right_child < @heap.size && @heap[right_child] < @heap[min_index]
min_index = right_child
end
return if min_index == index
swap(index, min_index)
sift_down(min_index)
end
def swap(i, j)
@heap[i], @heap[j] = @heap[j], @heap[i]
end
end
class Solution
def minimum_time(n, edges, disappear)
g = Array.new(n) { [] }
edges.each do |e|
x, y, wt = e
g[x] << [y, wt]
g[y] << [x, wt]
end
dis = Array.new(n, -1)
dis[0] = 0
pq = MinHeap.new
pq.push([0, 0])
until pq.empty?
dx, x = pq.pop
next if dx > dis[x]
g[x].each do |e|
y, wt = e
new_dis = dx + wt
if new_dis < disappear[y] && (dis[y] < 0 || new_dis < dis[y])
dis[y] = new_dis
pq.push([new_dis, y])
end
end
end
dis
end
end
Explanation:
The MinHeap class is implemented to manage the priority queue operations efficiently.
The push method adds a new element to the heap and ensures the heap property is maintained by sifting the element up.
The pop method removes and returns the minimum element from the heap, then ensures the heap property is maintained by sifting the element down.
The empty? method checks if the heap is empty.
In the Solution class, the MinHeap is used to manage the priority queue instead of sorting an array.
This should provide a more efficient implementation of the priority queue operations.
I:this code would be bring undefined method <=' for Array problem
chatgpt:
You're right, comparing arrays directly with the < or <= operators will lead to issues. We need to define the comparison logic explicitly for the elements stored in the heap. In this case, we need to compare the first elements of the arrays to maintain the priority based on distance.
Here’s the corrected implementation:
class MinHeap
def initialize
@heap = []
end
def push(val)
@heap << val
sift_up(@heap.size - 1)
end
def pop
return if @heap.empty?
swap(0, @heap.size - 1)
min = @heap.pop
sift_down(0)
min
end
def empty?
@heap.empty?
end
private
def sift_up(index)
parent = (index - 1) / 2
return if index <= 0 || compare(@heap[parent], @heap[index])
swap(index, parent)
sift_up(parent)
end
def sift_down(index)
left_child = 2 * index + 1
right_child = 2 * index + 2
min_index = index
if left_child < @heap.size && compare(@heap[left_child], @heap[min_index])
min_index = left_child
end
if right_child < @heap.size && compare(@heap[right_child], @heap[min_index])
min_index = right_child
end
return if min_index == index
swap(index, min_index)
sift_down(min_index)
end
def swap(i, j)
@heap[i], @heap[j] = @heap[j], @heap[i]
end
def compare(a, b)
a[0] <= b[0]
end
end
class Solution
def minimum_time(n, edges, disappear)
g = Array.new(n) { [] }
edges.each do |e|
x, y, wt = e
g[x] << [y, wt]
g[y] << [x, wt]
end
dis = Array.new(n, -1)
dis[0] = 0
pq = MinHeap.new
pq.push([0, 0])
until pq.empty?
dx, x = pq.pop
next if dx > dis[x]
g[x].each do |e|
y, wt = e
new_dis = dx + wt
if new_dis < disappear[y] && (dis[y] < 0 || new_dis < dis[y])
dis[y] = new_dis
pq.push([new_dis, y])
end
end
end
dis
end
end
Explanation:
The compare method in MinHeap explicitly compares the first elements of the arrays to maintain the priority based on distance.
The sift_up and sift_down methods use this compare method to ensure that the heap property is maintained correctly.
This avoids the issue of trying to use the < or <= operators directly on arrays.
最后顺利解决，代码通过测试
class MinHeap
def initialize
@heap = []
end
def push(val)
@heap << val
sift_up(@heap.size - 1)
end
def pop
return if @heap.empty?
swap(0, @heap.size - 1)
min = @heap.pop
sift_down(0)
min
end
def empty?
@heap.empty?
end
private
def sift_up(index)
parent = (index - 1) / 2
return if index <= 0 || compare(@heap[parent], @heap[index])
swap(index, parent)
sift_up(parent)
end
def sift_down(index)
left_child = 2 * index + 1
right_child = 2 * index + 2
min_index = index
if left_child < @heap.size && compare(@heap[left_child], @heap[min_index])
min_index = left_child
end
if right_child < @heap.size && compare(@heap[right_child], @heap[min_index])
min_index = right_child
end
return if min_index == index
swap(index, min_index)
sift_down(min_index)
end
def swap(i, j)
@heap[i], @heap[j] = @heap[j], @heap[i]
end
def compare(a, b)
a[0] <= b[0]
end
end
def minimum_time(n, edges, disappear)
g = Array.new(n) { [] }
edges.each do |e|
x, y, wt = e
g[x] << [y, wt]
g[y] << [x, wt]
end
dis = Array.new(n, -1)
dis[0] = 0
pq = MinHeap.new
pq.push([0, 0])
until pq.empty?
dx, x = pq.pop
next if dx > dis[x]
g[x].each do |e|
y, wt = e
new_dis = dx + wt
if new_dis < disappear[y] && (dis[y] < 0 || new_dis < dis[y])
dis[y] = new_dis
pq.push([new_dis, y])
end
end
end
dis
end
