Hdu-4661 Message Passing(树形DP)

There are n people numbered from 1 to n. Each people have a unique message. Some pairs of people can send messages directly to each other, and this relationship forms a structure of a tree. In one turn, exactly one person sends all messages s/he currently has to another person. What is the minimum number of turns needed so that everyone has all the messages? 
This is not your task. Your task is: count the number of ways that minimizes the number of turns. Two ways are different if there exists some k such that in the k-th turn, the sender or receiver is different in the two ways. 

Input

First line, number of test cases, T. 
Following are T test cases. 
For each test case, the first line is number of people, n. Following are n-1 lines. Each line contains two numbers. 

Sum of all n <= 1000000.

Output

T lines, each line is answer to the corresponding test case. Since the answers may be very large, you should output them modulo 10 ⁹+7.

Sample Input

221 231 22 3

Sample Output

2
6
题意:n个人构成一棵树,每个人有一个独特的信息,每轮最多有一个人把他携带的所有信息传递给和他相邻的人,假如最多传k轮所有人都能得到所有信息,问方案数有多少.
分析:考虑树上的每条边,可以看作有两个方向,那么每条边最多传入一次传出一次,所以最少传递次数的上限为2*(n-1),然后我们可以轻易的构造出来这种方案,只要选一个点为根每次都先通过n-1把所有的信息都传入给他,再通过n-1次将这个信息传回所有点就可以了,然后我们考虑如何统计方案,考虑每个点都作为一次汇点,假如传入a的方案数为f(a),那么我们可以根据一一对应的法则证明从a传出去的方案数也为f(a),所以最后的答案就是sigma(f(a)^2),f(a)可以用树形dp求出来,具体就是先随便找一个点作为根,然后dfs这棵树并且求出来每个点的子树符合拓扑序的所有方案数,然后再dfs一次求出以每个点为汇点的方案数.
#include <bits/stdc++.h>#define INF 1000111111#define N 1000005#define MOD 1000000007using namespace std;typedef long long ll;int T,n,x,y;ll ans,gc[N],igc[N],Ans[N],Down[N],Size[N];vector<int> G[N];void exgcd(ll a,ll b,ll &g,ll &x,ll &y){    if(!b) g=a,x=1,y=0;    else    {        exgcd(b,a%b,g,y,x);        y-=a/b*x;    }}ll inv(ll a,ll n){    ll d,x,y;    exgcd(a,n,d,x,y);    return d == 1 ? (x+n)%n : -1;}ll c(int x,int y){    return (gc[x]*igc[y] % MOD)*igc[x-y] % MOD;}void dfs1(int u,int fa){    Size[u] = 1,Down[u] = 1;    for(int v : G[u])     if(v != fa)     {         dfs1(v,u);         Size[u] += Size[v];         if(!Down[u]) Down[u] = Down[v];         else Down[u] = (Down[u]*Down[v] % MOD)*c(Size[u] - 1,Size[v]) % MOD;     }}void dfs2(int u,int fa){    if(u == 1) Ans[u] = Down[u];    else    {        ll res = (Ans[fa]*inv(Down[u],MOD) % MOD)*inv(c(n-1,Size[u]),MOD) % MOD;        Ans[u] = (c(n-1,Size[u]-1)*Down[u] % MOD)*res % MOD;    }    ans = (ans + Ans[u]*Ans[u] % MOD) % MOD;    for(int v : G[u])     if(v != fa) dfs2(v,u);}int main(){    gc[0] = igc[0] = 1;    for(int i = 1;i < N;i++) gc[i] = (gc[i-1]*i) % MOD;    for(int i = 1;i < N;i++) igc[i] = inv(gc[i],MOD);    scanf("%d",&T);    while(T--)    {        ans = 0;        scanf("%d",&n);        for(int i = 1;i <= n;i++) G[i].clear();        for(int i = 1;i < n;i++)        {            scanf("%d%d",&x,&y);            G[x].push_back(y);            G[y].push_back(x);        }        dfs1(1,1);        dfs2(1,1);        cout<<ans<<endl;    }}