leetcode2065 最大化一张图中的路径价值

思路：

暴搜。

实现：

 1 class Solution
 2 {
 3 public:
 4     int res = 0; 
 5     vector<int> vis;
 6     void dfs(int x, int time, int val, vector<int>& values, vectorint, int>>>& g)
 7     {
 8         if (x == 0) res = max(res, val);
 9         int n = g[x].size();
10         int maxn = 0;
11         for (int i = 0; i < n; i++)
12         {
13             auto tmp = g[x][i];
14             int to = tmp.first, cost = tmp.second;
15             if (cost > time) continue;
16             if (!vis[to])
17             {
18                 vis[to] = true;
19                 dfs(to, time - cost, val + values[to], values, g);
20                 vis[to] = false;
21             }
22             else dfs(to, time - cost, val, values, g);
23         }
24     }
25     int maximalPathQuality(vector<int>& values, vectorint>>& edges, int maxTime)
26     {
27         res = values[0];
28         int n = values.size();
29         vis.resize(n, 0);
30         int m = edges.size();
31         vectorint, int>>> g(n, vectorint, int>>());
32         for (int i = 0; i < m; i++)
33         {
34             auto tmp = edges[i];
35             int a = tmp[0], b = tmp[1], c = tmp[2];
36             g[a].push_back({b, c});
37             g[b].push_back({a, c});
38         }
39         dfs(0, maxTime, 0, values, g);
40         return res;
41     }
42 };