牛客国庆集训派对Day3 Solution

A    Knight

留坑。

 

B    Tree

思路：两次树形DP，但是要考虑0没有逆元

可以用前缀后缀做

 

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3  
 4 #define N 1000010
 5 #define ll long long
 6  
 7 const ll MOD = (ll)1e9 + 7;
 8  
 9  
10 int n;
11 ll dp[N], dp2[N];
12 ll prefix[N], suffix[N];
13 vector <int> G[N]; 
14  
15 void Init()
16 {
17     for (int i = 1; i <= n; ++i) dp[i] = 1, dp2[i] = 0, G[i].clear();
18 }
19  
20 void DFS(int u, int fa)
21 {
22     for (auto v : G[u])
23     {
24         if (v == fa) continue;
25         DFS(v, u);
26         dp[u] = dp[u] * (dp[v] + 1) % MOD;
27     }
28 }
29  
30 void DFS2(int u, int fa)
31 {
32     prefix[0] = 1;
33     suffix[G[u].size() + 1] = 1;
34     int len = G[u].size();
35     for (int i = 0; i < len; ++i)
36     {
37         int v = G[u][i];
38         if (v == fa) prefix[i + 1] = prefix[i];
39         else prefix[i + 1] = prefix[i] * (dp[G[u][i]] + 1) % MOD;
40     }
41     for (int i = len - 1; i >= 0; --i)
42     {
43         int v = G[u][i];
44         if (v == fa) suffix[i + 1] = suffix[i + 2];
45         else suffix[i + 1] = suffix[i + 2] * (dp[G[u][i]] + 1) % MOD;
46  
47     }
48     for (int i = 0; i < len; ++i) 
49     {
50         int v = G[u][i];
51         if (v == fa) continue; 
52         dp2[v] = prefix[i] % MOD * suffix[i + 2] % MOD * (dp2[u] + 1) % MOD; 
53     }
54     for (auto v : G[u])
55     {
56         if (v == fa) continue;
57         DFS2(v, u);
58     }
59 }
60  
61 void Run()
62 {
63     while (scanf("%d", &n) != EOF)
64     {
65         Init();
66         for (int i = 1, u, v; i < n; ++i)
67         {
68             scanf("%d%d", &u, &v);
69             G[u].push_back(v);
70             G[v].push_back(u);
71         }
72         DFS(1, 0); DFS2(1, 0);
73         for (int i = 1; i <= n; ++i) printf("%lld\n", dp[i] * (dp2[i] + 1) % MOD);
74     }
75 }
76  
77 int main()
78 {
79     #ifdef LOCAL
80         freopen("Test.in", "r", stdin);
81     #endif
82  
83     Run();
84     return 0;
85 }

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