Leetcode 2858. Minimum Edge Reversals So Every Node Is Reachable

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2858. Minimum Edge Reversals So Every Node Is Reachable

2858. Minimum Edge Reversals So Every Node Is Reachable

讲解：
【模板】换根 DP（Python/Java/C++/Go/JS）

类似题目：
834. Sum of Distances in Tree
2581. Count Number of Possible Root Nodes

re-rooting/换根DP

一般这种题目是求如果以每个节点为根节点的情况下，达到要求的情况个数。一般思路是，先从以0为根节点出发，计算答案，然后从0深度遍历，在转换关系中以0计算出的答案为基础，计算出当前根节点的情况个数。所以，重点是如何在转换中寻找关系。多练习。

class Solution {
    public int[] minEdgeReversals(int n, int[][] edges) {
        Map<Integer, Set<int[]>> g = new HashMap<>();
        for (int[] edge : edges) {
            int u = edge[0], v = edge[1];
            g.putIfAbsent(u, new HashSet<>());
            g.putIfAbsent(v, new HashSet<>());
            g.get(u).add(new int[]{v, 1});
            g.get(v).add(new int[]{u, -1});
        }
        int[] ans = new int[n];
        
        dfs(0, -1, g, ans);
        
        reRoot(0, -1, g, ans);
        
        return ans;
    }
    
    private void dfs(int cur, int fa, Map<Integer, Set<int[]>> g, int[] ans) {
        for (int[] next : g.get(cur)) {
            if (next[0] == fa) {
                continue;
            }
            ans[0] += next[1] == -1 ? 1 : 0;
            dfs(next[0], cur, g, ans);
        }
    }
    
    private void reRoot(int cur, int fa, Map<Integer, Set<int[]>> g, int[] ans) {
        for (int[] next : g.get(cur)) {
            if (next[0] == fa) {
                continue;
            }
            ans[next[0]] = ans[cur] + next[1];
            reRoot(next[0], cur, g, ans);
        }
    }
}