851. Loud and Rich —— weekly contest 87

Posted 2020-11-17 jinjin-2018

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。

 1 void DFS(vector<int> &visited, vector<vector<int> >& G, int x,vector<int>& quiet, int &res,int &minid){
 2         visited[x] = 1;
 3         if(res > quiet[x]){
 4             res = quiet[x];
 5             minid = x;
 6         }
 7         for(auto i:G[x]){
 8             if(visited[i] == 0){
 9                 DFS(visited,G,i,quiet,res,minid);
10             }
11         }
12         return;
13     }
14     vector<int> loudAndRich(vector<vector<int>>& richer, vector<int>& quiet) {
15         vector<int> ans;
16         int n = quiet.size();
17         vector<vector<int> > G;
18         G.resize(n);
19         for(auto i : richer){
20             G[i[1]].push_back(i[0]);
21         }
22         vector<int> visited;
23         
24         for(int i = 0 ; i < n; i++){
25             int res = 100000;
26             int minid = -1;
27             visited.assign(n,0);
28             DFS(visited,G,i,quiet,res,minid);
29             ans.push_back(minid);
30         }
31         return ans;
32     }