HDU 4661 message passing (wood DP & amp; Combinatorics)

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Message Passing

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 1187 Accepted Submission(s): 423

Problem Description

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

    2
2
1 2
3
1 2
2 3

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Sample Output

    2
6

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Source

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The question ：

n personal , Form a tree relationship . Everyone has a unique message . Everyone can pass their information through the tree . Share with his neighbors , After sharing . The person who is shared has his original information and shared information .

Each share is an operation . Ask how many sharing methods are there for each person to have all their information for the minimum number of times .

Ideas ：

easy Out of . In the shortest time . Of course, each node wants to send its own information out once . And receive a message , That is, by the tree 2 times 2*(n-1). Then we can prove . In the shortest time , All delivery methods have one “ Information conversion point ”—— The information of other nodes is first transferred to this node , Then the information is transferred from this node to other nodes .

In fact, what we need is how many kinds of topological orders . Definition dp[u] Express u Below the node . to u How many kinds of topological orders of nodes ,cnt[u] Express u How many children and grandchildren are there ,f[i] = i!(i The factorial ).c[i][j] Represents the number of combinations .

If it has v1,v2,v3 Nodes . Their topological orders are dp[v1],dp[v2],dp[v3] So many .

that dp[u] = c[cnt[u]-1][cnt[v1]] * c[cnt[u]-1-cnt[v1]][cnt[v2]] * c[cnt[u]-1-cnt[v1]-cnt[v2]][cnt[v3]] * dp[v1] * dp[v2] * dp[v3]（ Push this by yourself ）. After simplification . obtain dp[u] = f[cnt[u]-1] / ( f[cnt[v1]] * f[cnt[v2]] * f[cnt[v3]] ) * dp[v1] * dp[v2] * dp[v3] . We can o（n） The time complexity is calculated with 1 All nodes with root dp value （ So in order to 1 The answer for root is calculated ）. And some other auxiliary information . Then traverse down according to the tree structure . Calculate the answers rooted in other nodes respectively . The above is the online thinking . What I want to say is my understanding . Why do you know the number of topological orders of each subtree . You can exit your topological order number . In fact, it's very easy to understand . When the topological order of each subtree is determined . When determining the overall order .

That is to get a length of cnt[u] The topological sequence .

For subtrees i.

There is also a length of cnt[i] The topological sequence , So it's about cnt[u] Look inside cnt[i] A place . Look for other subtrees in the remaining subtrees .

And how to deduce when changing the root . Write first

dp[u]’=dp[u]*(n-sz[v]-1)!*sz[v]!/((n-1)!*dp[v])

dp[v]’=dp[v]*(n-1)!*dp[u]’/((sz[v]-1)!*(n-sz[v])!).

Into the dp[u]’ You can ask for a lot of things . So don't rush to get the final answer when pushing the formula . There is also why the answer number is the square of the topological ordinal .

Because when the information is sent back, it is your topological order . Same as topological ordinal .

Each positive topological order can be combined with an inverse topological order . So there are square species .

See code for details ：

#include<algorithm>
#include<iostream>
#include<string.h>
#include<stdio.h>
using namespace std;
const int INF=0x3f3f3f3f;
const int maxn=1000010;
#pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long ll;
const ll mod=1e9+7;
ll fac[maxn],dp[maxn],ans;
int cnt,sz[maxn],n;
struct node
{
    int v;
    node *next;
} ed[maxn<<1],*head[maxn];
void adde(int u,int v)
{
    ed[cnt].v=v;
    ed[cnt].next=head[u];
    head[u]=&ed[cnt++];
}
ll pow_mod(ll x,int k)
{
    ll base=x,ret=1;
    while(k)
    {
        if(k&1)
            ret=(ret*base)%mod;
        base=(base*base)%mod;
        k>>=1;
    }
    return ret;
}
ll ni(ll x){    return pow_mod(x,mod-2);    }
void dfs(int fa,int u)
{
    ll tp;
    dp[u]=tp=sz[u]=1;
    for(node *p=head[u];p!=NULL;p=p->next)
    {
        int v=p->v;
        if(v==fa)
            continue;
        dfs(u,v);
        tp=(tp*ni(fac[sz[v]]))%mod;
        dp[u]=(dp[u]*dp[v])%mod;
        sz[u]+=sz[v];
    }
    dp[u]=(dp[u]*fac[sz[u]-1]%mod*tp)%mod;
}
void solve(int fa,int u,ll tp)
{
    ll tt;
    ans=(ans+tp*tp%mod)%mod;
    for(node *p=head[u];p!=NULL;p=p->next)
    {
        int v=p->v;
        if(v==fa)
            continue;
        tt=(tp*fac[n-sz[v]-1]%mod*fac[sz[v]])%mod;
        tt=(tt*ni(fac[sz[v]-1])%mod*ni(fac[n-sz[v]]))%mod;
        solve(u,v,tt);
    }
}
int main()
{
    int t,i,u,v,rt;

    fac[0]=fac[1]=1;
    for(i=2;i<maxn;i++)
        fac[i]=(i*fac[i-1])%mod;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        cnt=0,ans=0;
        for(i=1;i<=n;i++)
            head[i]=NULL;
        for(i=1;i<n;i++)
        {
            scanf("%d%d",&u,&v);
            adde(u,v);
            adde(v,u);
        }
        rt=(n+1)/2;
        dfs(-1,rt);
        solve(-1,rt,dp[rt]);
        printf("%I64d\n",ans);
    }
    return 0;
}

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