原题链接：Leetcode 2192. 有向无环图中一个节点的所有祖先

class Solution {
public:
    vector<vector<int>> adj;
    vector<vector<int>> res;
    vector<int> indegree;
    map<pair<int,int>,int> mp;
    void dfs(int now)
    {
        for(auto x:adj[now])
        {
            if(!mp[{now,x}])
            {
                res[now].push_back(x);
                mp[{now,x}]=1;
            }
            if(res[x].size()==0) dfs(x);
            for(auto y:res[x])
            {
                if(!mp[{now,y}])
                {
                    res[now].push_back(y);
                    mp[{now,y}]=1;
                }
            }
        }
    }
    vector<vector<int>> getAncestors(int n, vector<vector<int>>& edges) {
        res.resize(n);
        adj.resize(n);
        indegree.resize(n);
        for(auto x:edges)
        {
            int a=x[0],b=x[1];
            adj[b].push_back(a);
            indegree[a]++;
        }
        for(int i=0;i<n;i++)
        {
            if(!indegree[i])   dfs(i);
        }
        for(int i=0;i<n;i++) sort(res[i].begin(),res[i].end());
        return res;
    }
};

拓扑排序

class Solution {
public:
    vector<vector<int>> getAncestors(int n, vector<vector<int>>& edges) {
        vector<vector<int>> adj(n);
        vector<vector<int>> res(n);
        vector<int> indegree(n);
        vector<set<int>> pre(n);
        for(auto x:edges)
        {
            int a=x[0],b=x[1];
            adj[a].push_back(b);
            indegree[b]++;
        }
        queue<int> q; 
        for(int i=0;i<n;i++)
        {
            if(!indegree[i]) q.push(i);
        }
        while(!q.empty())
        {
            int now=q.front(); q.pop();
            for(auto x:adj[now])
            {
                indegree[x]--;
                pre[x].insert(pre[now].begin(),pre[now].end());
                pre[x].insert(now);
                if(!indegree[x]) q.push(x);
            }
        }
        for(int i=0;i<n;i++)
        {
            res[i].insert(res[i].end(),pre[i].begin(),pre[i].end());
        }
        return res;
    }
};