蓝桥杯 结点选择

思路：

无根树转有根树 + 树形dp。

实现：

 1 #include 
 2 #include 
 3 #include 
 4 #include 
 5 #include 
 6 using namespace std;
 7 
 8 const int MAXN = 100000;
 9 const int INF = 0x3f3f3f3f;
10 
11 int n, x, y;
12 int dp[MAXN + 5][2], a[MAXN + 5], f[MAXN + 5], vis[MAXN + 5];
13 vector<int> G[MAXN + 5];
14 
15 void trans(int root, int p)
16 {
17     for (int i = 0; i < G[root].size(); i++)
18     {
19         int tmp = G[root][i];
20         if (tmp != p)
21             trans(tmp, f[tmp] = root);
22     }
23 }
24 
25 void build()
26 {
27     for (int i = 1; i <= n; i++)
28         G[i].clear();
29     for (int i = 2; i <= n; i++)
30     {
31         G[f[i]].push_back(i);
32     }
33 }
34 
35 void dfs(int now)
36 {
37     vis[now] = 1;
38     for (int i = 0; i < G[now].size(); i++)
39     {
40         int tmp = G[now][i];
41         if (!vis[tmp])
42         {
43             dfs(tmp);
44         }
45         dp[now][1] += dp[tmp][0];
46         dp[now][0] += max(dp[tmp][1], dp[tmp][0]);
47     }
48     dp[now][1] += a[now];
49 }
50 
51 int main()
52 {
53     cin >> n;
54     for (int i = 1; i <= n; i++)
55     {
56         scanf("%d", &a[i]);
57     }
58     for (int i = 0; i < n - 1; i++)
59     {
60         scanf("%d %d", &x, &y);
61         G[x].push_back(y);
62         G[y].push_back(x);
63     }
64     int ans = -INF;
65     trans(1, -1);
66     build();
67     memset(dp, 0, sizeof(dp));
68     memset(vis, 0, sizeof(vis));
69     dfs(1);
70     ans = max(ans, max(dp[1][0], dp[1][1]));
71     printf("%d\n", ans);
72     return 0;
73 }

 

转载于:https://www.cnblogs.com/wangyiming/p/6657230.html