Description
An undirected graph of n nodes is defined by edgeList, where edgeList[i] = [ui, vi, disi] denotes an edge between nodes ui and vi with distance disi. Note that there may be multiple edges between two nodes, and the graph may not be connected.
Implement the DistanceLimitedPathsExist class:
DistanceLimitedPathsExist(int n, int[][] edgeList) Initializes the class with an undirected graph.boolean query(int p, int q, int limit) Returns true if there exists a path from p to q such that each edge on the path has a distance strictly less than limit, and otherwise false.
Example 1:
Input
["DistanceLimitedPathsExist", "query", "query", "query", "query"]
[[6, [[0, 2, 4], [0, 3, 2], [1, 2, 3], [2, 3, 1], [4, 5, 5]]], [2, 3, 2], [1, 3, 3], [2, 0, 3], [0, 5, 6]]
Output
[null, true, false, true, false]
Explanation
DistanceLimitedPathsExist distanceLimitedPathsExist = new DistanceLimitedPathsExist(6, [[0, 2, 4], [0, 3, 2], [1, 2, 3], [2, 3, 1], [4, 5, 5]]);
distanceLimitedPathsExist.query(2, 3, 2); // return true. There is an edge from 2 to 3 of distance 1, which is less than 2.
distanceLimitedPathsExist.query(1, 3, 3); // return false. There is no way to go from 1 to 3 with distances strictly less than 3.
distanceLimitedPathsExist.query(2, 0, 3); // return true. There is a way to go from 2 to 0 with distance < 3: travel from 2 to 3 to 0.
distanceLimitedPathsExist.query(0, 5, 6); // return false. There are no paths from 0 to 5.
Constraints:
2 <= n <= 1040 <= edgeList.length <= 104edgeList[i].length == 30 <= ui, vi, p, q <= n-1ui != vip != q1 <= disi, limit <= 109- At most
104 calls will be made to query.
Solutions
Solution 1
PythonJavaC++GoTypeScript
class PersistentUnionFind:
def __init__(self, n):
self.rank = [0] * n
self.p = list(range(n))
self.version = [inf] * n
def find(self, x, t=inf):
if self.p[x] == x or self.version[x] >= t:
return x
return self.find(self.p[x], t)
def union(self, a, b, t):
pa, pb = self.find(a), self.find(b)
if pa == pb:
return False
if self.rank[pa] > self.rank[pb]:
self.version[pb] = t
self.p[pb] = pa
else:
self.version[pa] = t
self.p[pa] = pb
if self.rank[pa] == self.rank[pb]:
self.rank[pb] += 1
return True
class DistanceLimitedPathsExist:
def __init__(self, n: int, edgeList: List[List[int]]):
self.puf = PersistentUnionFind(n)
edgeList.sort(key=lambda x: x[2])
for u, v, dis in edgeList:
self.puf.union(u, v, dis)
def query(self, p: int, q: int, limit: int) -> bool:
return self.puf.find(p, limit) == self.puf.find(q, limit)(code-box)
class PersistentUnionFind {
private final int inf = 1 << 30;
private int[] rank;
private int[] parent;
private int[] version;
public PersistentUnionFind(int n) {
rank = new int[n];
parent = new int[n];
version = new int[n];
for (int i = 0; i < n; i++) {
parent[i] = i;
version[i] = inf;
}
}
public int find(int x, int t) {
if (parent[x] == x || version[x] >= t) {
return x;
}
return find(parent[x], t);
}
public boolean union(int a, int b, int t) {
int pa = find(a, inf);
int pb = find(b, inf);
if (pa == pb) {
return false;
}
if (rank[pa] > rank[pb]) {
version[pb] = t;
parent[pb] = pa;
} else {
version[pa] = t;
parent[pa] = pb;
if (rank[pa] == rank[pb]) {
rank[pb]++;
}
}
return true;
}
}
public class DistanceLimitedPathsExist {
private PersistentUnionFind puf;
public DistanceLimitedPathsExist(int n, int[][] edgeList) {
puf = new PersistentUnionFind(n);
Arrays.sort(edgeList, (a, b) -> a[2] - b[2]);
for (var e : edgeList) {
puf.union(e[0], e[1], e[2]);
}
}
public boolean query(int p, int q, int limit) {
return puf.find(p, limit) == puf.find(q, limit);
}
}
/**
* Your DistanceLimitedPathsExist object will be instantiated and called as such:
* DistanceLimitedPathsExist obj = new DistanceLimitedPathsExist(n, edgeList);
* boolean param_1 = obj.query(p,q,limit);
*/(code-box)
class PersistentUnionFind {
private:
vector<int> rank;
vector<int> parent;
vector<int> version;
public:
PersistentUnionFind(int n)
: rank(n, 0)
, parent(n)
, version(n, INT_MAX) {
for (int i = 0; i < n; i++) {
parent[i] = i;
}
}
int find(int x, int t) {
if (parent[x] == x || version[x] >= t) {
return x;
}
return find(parent[x], t);
}
bool unionSet(int a, int b, int t) {
int pa = find(a, INT_MAX);
int pb = find(b, INT_MAX);
if (pa == pb) {
return false;
}
if (rank[pa] > rank[pb]) {
version[pb] = t;
parent[pb] = pa;
} else {
version[pa] = t;
parent[pa] = pb;
if (rank[pa] == rank[pb]) {
rank[pb]++;
}
}
return true;
}
};
class DistanceLimitedPathsExist {
private:
PersistentUnionFind puf;
public:
DistanceLimitedPathsExist(int n, vector<vector<int>>& edgeList)
: puf(n) {
sort(edgeList.begin(), edgeList.end(),
[](const vector<int>& a, const vector<int>& b) {
return a[2] < b[2];
});
for (const auto& edge : edgeList) {
puf.unionSet(edge[0], edge[1], edge[2]);
}
}
bool query(int p, int q, int limit) {
return puf.find(p, limit) == puf.find(q, limit);
}
};
/**
* Your DistanceLimitedPathsExist object will be instantiated and called as such:
* DistanceLimitedPathsExist* obj = new DistanceLimitedPathsExist(n, edgeList);
* bool param_1 = obj->query(p,q,limit);
*/(code-box)
type PersistentUnionFind struct {
rank []int
parent []int
version []int
}
func NewPersistentUnionFind(n int) *PersistentUnionFind {
rank := make([]int, n)
parent := make([]int, n)
version := make([]int, n)
for i := 0; i < n; i++ {
parent[i] = i
}
return &PersistentUnionFind{
rank: rank,
parent: parent,
version: version,
}
}
func (uf *PersistentUnionFind) find(x int, t int) int {
if uf.parent[x] == x || uf.version[x] >= t {
return x
}
return uf.find(uf.parent[x], t)
}
func (uf *PersistentUnionFind) union(a, b, t int) bool {
pa := uf.find(a, int(^uint(0)>>1))
pb := uf.find(b, int(^uint(0)>>1))
if pa == pb {
return false
}
if uf.rank[pa] > uf.rank[pb] {
uf.version[pb] = t
uf.parent[pb] = pa
} else {
uf.version[pa] = t
uf.parent[pa] = pb
if uf.rank[pa] == uf.rank[pb] {
uf.rank[pb]++
}
}
return true
}
type DistanceLimitedPathsExist struct {
puf *PersistentUnionFind
}
func Constructor(n int, edgeList [][]int) DistanceLimitedPathsExist {
sort.Slice(edgeList, func(i, j int) bool {
return edgeList[i][2] < edgeList[j][2]
})
puf := NewPersistentUnionFind(n)
for _, edge := range edgeList {
puf.union(edge[0], edge[1], edge[2])
}
return DistanceLimitedPathsExist{
puf: puf,
}
}
func (dle *DistanceLimitedPathsExist) Query(p, q, limit int) bool {
return dle.puf.find(p, limit) == dle.puf.find(q, limit)
}
/**
* Your DistanceLimitedPathsExist object will be instantiated and called as such:
* obj := Constructor(n, edgeList);
* param_1 := obj.Query(p,q,limit);
*/(code-box)
class PersistentUnionFind {
private rank: number[];
private parent: number[];
private version: number[];
constructor(n: number) {
this.rank = Array(n).fill(0);
this.parent = Array.from({ length: n }, (_, i) => i);
this.version = Array(n).fill(Infinity);
}
find(x: number, t: number): number {
if (this.parent[x] === x || this.version[x] >= t) {
return x;
}
return this.find(this.parent[x], t);
}
union(a: number, b: number, t: number): boolean {
const pa = this.find(a, Infinity);
const pb = this.find(b, Infinity);
if (pa === pb) {
return false;
}
if (this.rank[pa] > this.rank[pb]) {
this.version[pb] = t;
this.parent[pb] = pa;
} else {
this.version[pa] = t;
this.parent[pa] = pb;
if (this.rank[pa] === this.rank[pb]) {
this.rank[pb]++;
}
}
return true;
}
}
class DistanceLimitedPathsExist {
private puf: PersistentUnionFind;
constructor(n: number, edgeList: number[][]) {
this.puf = new PersistentUnionFind(n);
edgeList.sort((a, b) => a[2] - b[2]);
for (const [u, v, dis] of edgeList) {
this.puf.union(u, v, dis);
}
}
query(p: number, q: number, limit: number): boolean {
return this.puf.find(p, limit) === this.puf.find(q, limit);
}
}
/**
* Your DistanceLimitedPathsExist object will be instantiated and called as such:
* var obj = new DistanceLimitedPathsExist(n, edgeList)
* var param_1 = obj.query(p,q,limit)
*/(code-box)
