https://leetcode.cn/problems/most-profitable-path-in-a-tree

Author: masx200

README.md

@@ -127,6 +127,8 @@ https://leetcode.cn/problems/design-movie-rental-system/
 
 https://leetcode.cn/problems/XxZZjK/
 
+https://leetcode.cn/problems/most-profitable-path-in-a-tree
+
 https://leetcode.cn/problems/magical-string/
 
 https://leetcode.cn/problems/er-cha-sou-suo-shu-de-hou-xu-bian-li-xu-lie-lcof/

most-profitable-path-in-a-tree/index.ts

@@ -0,0 +1,53 @@
+export default function mostProfitablePath(
+    edges: number[][],
+    bob: number,
+    amount: number[],
+): number {
+    const g: number[][] = amount.map(() => []);
+
+    for (const [x, y] of edges) {
+        g[x].push(y);
+        g[y].push(x);
+    }
+    const bobTime = amount.map(() => Infinity);
+    function dfsBob(x: number, fa: number, t: number): boolean {
+        if (x === 0) {
+            bobTime[x] = t;
+            return true;
+        }
+        let found0 = false;
+
+        for (const y of g[x]) {
+            if (y != fa && dfsBob(y, x, t + 1)) {
+                found0 = true;
+            }
+        }
+        if (found0) {
+            bobTime[x] = t; // 只有可以到达 0 才标记访问时间
+        }
+        return found0;
+    }
+    dfsBob(bob, -1, 0);
+    g[0].push(-1);
+    let ans = -Infinity;
+
+    function dfsAlice(x: number, fa: number, aliceTime: number, sum: number) {
+        if (aliceTime < bobTime[x]) {
+            sum += amount[x];
+        } else if (aliceTime == bobTime[x]) {
+            sum += amount[x] / 2;
+        }
+        if (g[x].length == 1) {
+            // 叶子
+            ans = Math.max(ans, sum); // 更新答案
+            return;
+        }
+        for (const y of g[x]) {
+            if (y != fa) {
+                dfsAlice(y, x, aliceTime + 1, sum);
+            }
+        }
+    }
+    dfsAlice(0, -1, 0, 0);
+    return ans;
+}