[2019 J Xuzhou network game title] Random Access Iterator

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Roughly meaning of the questions: starting from the root node, the node x has a son [x] into the inferior probability son node, find the probability of reaching the deepest depth. Son number of son nodes [x] is the x node.

And yet they do not any out of the mentality of collapse.

The official looked down and found himself the general idea of ​​the solution to a problem is not wrong, but the details are too weak Orz.

The general idea: DP set [x] as the root node of the x, find the probability of reaching the deepest depth. Xianpao again dfs, find the number of children for each point, each point of the depth and the deepest depths.

We can find a point x does not reach the deepest depth of the probability of $ cnt = 1 - (\ tfrac {\ sum_ {y \ epsilon x} dp [y]} {son [x]}) $

The $ dp [x] = 1-cnt ^ {son [x]} $ is the son [x] times the probability to reach the deepest depth.

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 const int maxn = 2e6 + 10;
 5 const ll mod = 1e9 + 7;
 6 struct node {
 7     int s, e, next;
 8 }edge[maxn];
 9 int head[maxn], len;
10 void init() {
11     memset(head, -1, sizeof(head));
12     len = 0;
13 }
14 void add(int s, int e) {
15     edge[len].e = e;
16     edge[len].next = head[s];
17     head[s] = len++;
18 }
19 int mxd[maxn], d[maxn], son[maxn];
20 ll dp[maxn];
21 ll qpow(ll a, ll b, ll mod) {
22     ll ans = 1;
23     while (b) {
24         if (b & 1)ans = ans * a % mod;
25         a = a * a % mod;
26         b >>= 1;
27     }
28     return ans;
29 }
30 void dfs(int x, int fa) {
31     mxd[x] = d[x];
32     for (int i = head[x]; i != -1; i = edge[i].next) {
33         int y = edge[i].e;
34         if (y == fa)continue;
35         d[y] = d[x] + 1;
36         son[x]++;
37         dfs(y, x);
38         mxd[x] = max(mxd[x], mxd[y]);
39     }
40 }
41 void dfs1(int x, int fa) {
42     ll sum = 0;
43     for (int i = head[x]; i != -1; i = edge[i].next) {
44         int y = edge[i].e;
45         if (y == fa)continue;
46         if (mxd[y] == mxd[1]) {
47             dfs1(y, x);
48             sum = (sum + dp[y]) % mod;
49         }
50     }
51     if (son[x] == 0)
52         dp[x] = 1;
53     else {
54         ll p = sum * qpow(son[x], mod - 2, mod) % mod;
55         p = (1 - p + mod) % mod;
56         dp[x] = (1 - qpow(p, son[x], mod) + mod) % mod;
57     }
58 }
59 int main() {
60     int n;
61     scanf("%d", &n);
62     init();
63     for (int i = 1; i < n; i++) {
64         int x, y;
65         scanf("%d%d", &x, &y);
66         add(x, y), add(y, x);
67     }
68     dfs(1, 0);
69     dfs1(1, 0);
70     printf("%lld\n", dp[1]);
71 }