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feat: add solutions to lc problem: No.1697.Checking Exsitence of Edge Length Limited Paths
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Diff for: ‎solution/1600-1699/1697.Checking Existence of Edge Length Limited Paths/README.md

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Original file line numberDiff line numberDiff line change
@@ -48,27 +48,230 @@
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<li>两个点之间可能有 <strong>多条</strong> 边。</li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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并查集。
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模板 1——朴素并查集：
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```python
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# 初始化，p存储每个点的父节点
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p = list(range(n))
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# 返回x的祖宗节点
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def find(x):
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if p[x] != x:
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# 路径压缩
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p[x] = find(p[x])
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return p[x]
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# 合并a和b所在的两个集合
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p[find(a)] = find(b)
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```
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模板 2——维护 size 的并查集：
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```python
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# 初始化，p存储每个点的父节点，size只有当节点是祖宗节点时才有意义，表示祖宗节点所在集合中，点的数量
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p = list(range(n))
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size = [1] * n
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# 返回x的祖宗节点
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def find(x):
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if p[x] != x:
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# 路径压缩
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p[x] = find(p[x])
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return p[x]
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# 合并a和b所在的两个集合
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if find(a) != find(b):
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size[find(b)] += size[find(a)]
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p[find(a)] = find(b)
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```
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模板 3——维护到祖宗节点距离的并查集：
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```python
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# 初始化，p存储每个点的父节点，d[x]存储x到p[x]的距离
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p = list(range(n))
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d = [0] * n
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# 返回x的祖宗节点
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def find(x):
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if p[x] != x:
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t = find(p[x])
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d[x] += d[p[x]]
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p[x] = t
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return p[x]
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# 合并a和b所在的两个集合
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p[find(a)] = find(b)
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d[find(a)] = distance
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```
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对于本题，可以转换为：将小于 limit 的所有边加入图中，判断此时是否存在从 pj 到 qj 的一条路径。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def distanceLimitedPathsExist(self, n: int, edgeList: List[List[int]], queries: List[List[int]]) -> List[bool]:
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p = list(range(n))
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def find(x):
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if p[x] != x:
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p[x] = find(p[x])
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return p[x]
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edgeList.sort(key=lambda x: x[2])
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m = len(queries)
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indexes = list(range(m))
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indexes.sort(key=lambda x: queries[x][2])
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ans = [False] * m
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i = 0
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for j in indexes:
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pj, qj, limit = queries[j][0], queries[j][1], queries[j][2]
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while i < len(edgeList) and edgeList[i][2] < limit:
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u, v = edgeList[i][0], edgeList[i][1]
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p[find(u)] = find(v)
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i += 1
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ans[j] = find(pj) == find(qj)
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return ans
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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private int[] p;
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public boolean[] distanceLimitedPathsExist(int n, int[][] edgeList, int[][] queries) {
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p = new int[n];
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for (int i = 0; i < n; ++i) {
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p[i] = i;
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}
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int m = queries.length;
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Integer[] indexes = new Integer[m];
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for (int i = 0; i < m; ++i) {
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indexes[i] = i;
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}
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Arrays.sort(indexes, Comparator.comparingInt(i -> queries[i][2]));
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Arrays.sort(edgeList, Comparator.comparingInt(a -> a[2]));
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boolean[] ans = new boolean[m];
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int i = 0;
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for (int j : indexes) {
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int pj = queries[j][0], qj = queries[j][1], limit = queries[j][2];
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while (i < edgeList.length && edgeList[i][2] < limit) {
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int u = edgeList[i][0], v = edgeList[i][1];
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p[find(u)] = find(v);
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++i;
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}
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ans[j] = find(pj) == find(qj);
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}
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return ans;
180+
}
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182+
private int find(int x) {
183+
if (p[x] != x) {
184+
p[x] = find(p[x]);
185+
}
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return p[x];
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}
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}
189+
```
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### **C++**
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```cpp
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class Solution {
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public:
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vector<int> p;
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vector<bool> distanceLimitedPathsExist(int n, vector<vector<int>>& edgeList, vector<vector<int>>& queries) {
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p.resize(n);
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for (int i = 0; i < n; ++i) p[i] = i;
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sort(edgeList.begin(), edgeList.end(), [](const auto& e1, const auto& e2) {
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return e1[2] < e2[2];
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});
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int m = queries.size();
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vector<int> indexes(m);
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for (int i = 0; i < m; ++i) indexes[i] = i;
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sort(indexes.begin(), indexes.end(), [&](int i, int j) {
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return queries[i][2] < queries[j][2];
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});
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vector<bool> ans(m, false);
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int i = 0;
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for (int j : indexes)
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{
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int pj = queries[j][0], qj = queries[j][1], limit = queries[j][2];
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while (i < edgeList.size() && edgeList[i][2] < limit)
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{
218+
int u = edgeList[i][0], v = edgeList[i][1];
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p[find(u)] = find(v);
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++i;
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}
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ans[j] = find(pj) == find(qj);
223+
}
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return ans;
225+
}
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int find(int x) {
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if (p[x] != x) p[x] = find(p[x]);
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return p[x];
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}
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};
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```
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### **Go**
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```go
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var p []int
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func distanceLimitedPathsExist(n int, edgeList [][]int, queries [][]int) []bool {
240+
p = make([]int, n)
241+
for i := 0; i < n; i++ {
242+
p[i] = i
243+
}
244+
sort.Slice(edgeList, func(i, j int) bool {
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return edgeList[i][2] < edgeList[j][2]
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})
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m := len(queries)
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indexes := make([]int, m)
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for i := 0; i < m; i++ {
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indexes[i] = i
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}
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sort.Slice(indexes, func(i, j int) bool {
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return queries[indexes[i]][2] < queries[indexes[j]][2]
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})
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ans := make([]bool, m)
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i := 0
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for _, j := range indexes {
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pj, qj, limit := queries[j][0], queries[j][1], queries[j][2]
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for i < len(edgeList) && edgeList[i][2] < limit {
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u, v := edgeList[i][0], edgeList[i][1]
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p[find(u)] = find(v)
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i++
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}
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ans[j] = find(pj) == find(qj)
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}
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return ans
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}
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func find(x int) int {
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if p[x] != x {
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p[x] = find(p[x])
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}
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return p[x]
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}
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```
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### **...**

Diff for: ‎solution/1600-1699/1697.Checking Existence of Edge Length Limited Paths/README_EN.md

+144-2
Original file line numberDiff line numberDiff line change
@@ -44,21 +44,163 @@ For the second query, there is a path (0 -&gt; 1 -&gt; 2) of two edges with dist
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<li>There may be <strong>multiple</strong> edges between two nodes.</li>
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</ul>
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## Solutions
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<!-- tabs:start -->
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### **Python3**
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```python
55-
54+
class Solution:
55+
def distanceLimitedPathsExist(self, n: int, edgeList: List[List[int]], queries: List[List[int]]) -> List[bool]:
56+
p = list(range(n))
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58+
def find(x):
59+
if p[x] != x:
60+
p[x] = find(p[x])
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return p[x]
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63+
edgeList.sort(key=lambda x: x[2])
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m = len(queries)
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indexes = list(range(m))
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indexes.sort(key=lambda x: queries[x][2])
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ans = [False] * m
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i = 0
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for j in indexes:
70+
pj, qj, limit = queries[j][0], queries[j][1], queries[j][2]
71+
while i < len(edgeList) and edgeList[i][2] < limit:
72+
u, v = edgeList[i][0], edgeList[i][1]
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p[find(u)] = find(v)
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i += 1
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ans[j] = find(pj) == find(qj)
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return ans
5677
```
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### **Java**
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```java
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class Solution {
83+
private int[] p;
84+
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public boolean[] distanceLimitedPathsExist(int n, int[][] edgeList, int[][] queries) {
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p = new int[n];
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for (int i = 0; i < n; ++i) {
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p[i] = i;
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}
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int m = queries.length;
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Integer[] indexes = new Integer[m];
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for (int i = 0; i < m; ++i) {
93+
indexes[i] = i;
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}
95+
Arrays.sort(indexes, Comparator.comparingInt(i -> queries[i][2]));
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Arrays.sort(edgeList, Comparator.comparingInt(a -> a[2]));
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boolean[] ans = new boolean[m];
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int i = 0;
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for (int j : indexes) {
100+
int pj = queries[j][0], qj = queries[j][1], limit = queries[j][2];
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while (i < edgeList.length && edgeList[i][2] < limit) {
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int u = edgeList[i][0], v = edgeList[i][1];
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p[find(u)] = find(v);
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++i;
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}
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ans[j] = find(pj) == find(qj);
107+
}
108+
return ans;
109+
}
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private int find(int x) {
112+
if (p[x] != x) {
113+
p[x] = find(p[x]);
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}
115+
return p[x];
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}
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}
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```
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### **C++**
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122+
```cpp
123+
class Solution {
124+
public:
125+
vector<int> p;
126+
127+
vector<bool> distanceLimitedPathsExist(int n, vector<vector<int>>& edgeList, vector<vector<int>>& queries) {
128+
p.resize(n);
129+
for (int i = 0; i < n; ++i) p[i] = i;
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sort(edgeList.begin(), edgeList.end(), [](const auto& e1, const auto& e2) {
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return e1[2] < e2[2];
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});
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int m = queries.size();
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vector<int> indexes(m);
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for (int i = 0; i < m; ++i) indexes[i] = i;
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sort(indexes.begin(), indexes.end(), [&](int i, int j) {
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return queries[i][2] < queries[j][2];
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});
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vector<bool> ans(m, false);
141+
int i = 0;
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for (int j : indexes)
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{
144+
int pj = queries[j][0], qj = queries[j][1], limit = queries[j][2];
145+
while (i < edgeList.size() && edgeList[i][2] < limit)
146+
{
147+
int u = edgeList[i][0], v = edgeList[i][1];
148+
p[find(u)] = find(v);
149+
++i;
150+
}
151+
ans[j] = find(pj) == find(qj);
152+
}
153+
return ans;
154+
}
155+
156+
int find(int x) {
157+
if (p[x] != x) p[x] = find(p[x]);
158+
return p[x];
159+
}
160+
};
161+
```
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### **Go**
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```go
166+
var p []int
167+
168+
func distanceLimitedPathsExist(n int, edgeList [][]int, queries [][]int) []bool {
169+
p = make([]int, n)
170+
for i := 0; i < n; i++ {
171+
p[i] = i
172+
}
173+
sort.Slice(edgeList, func(i, j int) bool {
174+
return edgeList[i][2] < edgeList[j][2]
175+
})
176+
m := len(queries)
177+
indexes := make([]int, m)
178+
for i := 0; i < m; i++ {
179+
indexes[i] = i
180+
}
181+
sort.Slice(indexes, func(i, j int) bool {
182+
return queries[indexes[i]][2] < queries[indexes[j]][2]
183+
})
184+
ans := make([]bool, m)
185+
i := 0
186+
for _, j := range indexes {
187+
pj, qj, limit := queries[j][0], queries[j][1], queries[j][2]
188+
for i < len(edgeList) && edgeList[i][2] < limit {
189+
u, v := edgeList[i][0], edgeList[i][1]
190+
p[find(u)] = find(v)
191+
i++
192+
}
193+
ans[j] = find(pj) == find(qj)
194+
}
195+
return ans
196+
}
197+
198+
func find(x int) int {
199+
if p[x] != x {
200+
p[x] = find(p[x])
201+
}
202+
return p[x]
203+
}
62204
```
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### **...**

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