851. Loud and Rich —— weekly contest 87

851. Loud and Rich

题目链接：https://leetcode.com/problems/loud-and-rich/description/

思路：有向图DFS,记录最小的quiet值

注意点：可优化，记忆性搜索，每次搜到已经记录过的值时可直接比较不需要进一步搜下去。

void DFS(vector<int> &visited, vector<vector<int> >& G, int x,vector<int>& quiet, int &res,int &minid){
        visited[x] = 1;
        if(res > quiet[x]){
            res = quiet[x];
            minid = x;
        }
        for(auto i:G[x]){
            if(visited[i] == 0){
                DFS(visited,G,i,quiet,res,minid);
            }
        }
        return;
    }
    vector<int> loudAndRich(vector<vector<int>>& richer, vector<int>& quiet) {
        vector<int> ans;
        int n = quiet.size();
        vector<vector<int> > G;
        G.resize(n);
        for(auto i : richer){
            G[i[1]].push_back(i[0]);
        }
        vector<int> visited;
        
        for(int i = 0 ; i < n; i++){
            int res = 100000;
            int minid = -1;
            visited.assign(n,0);
            DFS(visited,G,i,quiet,res,minid);
            ans.push_back(minid);
        }
        return ans;
    }