feat: 2076. Process Restricted Friend Requests : union find

Author: upupming

Diff for: leetcode/残酷刷题/2076. Process Restricted Friend Requests.js

@@ -0,0 +1,47 @@
+/**
+ * @param {number} n
+ * @param {number[][]} restrictions
+ * @param {number[][]} requests
+ * @return {boolean[]}
+ */
+ var friendRequests = function(n, restrictions, requests) {
+  /*
+  对于 [a, b] 一个 request 来说，如果 (x, y) 存在于 restrictions 里面
+  就说明不能够合在一起，x 是 a 代表的集合中的任意元素，y 是 b 代表的集合中的任意元素
+  因此 restrictions 也是需要去动态更新的，并查集更新的时候要更新成根节点的限制关系
+
+  0 x 1
+  1 - 2
+  0 x 2
+  */
+  const r = [...Array(n)].map(() => [])
+  for (const [x, y] of restrictions) {
+      r[x][y] = r[y][x] = 1
+  }
+  const fa = [...Array(n)].map((_, idx) => idx)
+  const get = (x) => {
+      if (fa[x] === x) return x
+      return fa[x] = get(fa[x])
+  }
+  function merge (x, y) {
+    const fx = get(x); const fy = get(y)
+    fa[fx] = fy
+  }
+  const m = requests.length
+  const ans = Array(m).fill(false)
+  for (let i = 0; i < m; i++) {
+      const [a, b] = requests[i]
+      if (r[get(a)][get(b)]) ans[i] = false
+      else {
+          ans[i] = true
+
+          for (let j = 0; j < n; j++) {
+              if (r[get(j)][get(a)]) r[get(j)][get(b)] = r[get(b)][get(j)] = true
+              if (r[get(j)][get(b)]) r[get(j)][get(a)] = r[get(a)][get(j)] = true
+          }
+
+          merge(a, b)
+      }
+  }
+  return ans
+};