HDOJ 5379 Mahjong tree 树形DP

在一棵树上给所有点标号，要求任意一个子树里的点编号连续，每一个点的儿子编号连续。 那么，一个点的非叶子儿子应该是连续的，即一个点的非叶子儿子最多只有两个。 对于每一个点，我们把它的叶子儿子的个数记作S，所有儿子的方案数积为T。当非叶子儿子节点个数小于2的时候，方案数为2T*(S!). 当非叶子儿子节点数等于2的时候，这个点为根的子树合法方案数位T*（S!）. 这样dfs一遍即可以处理整棵树的方案数。

Mahjong tree

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 742    Accepted Submission(s): 227

Problem Description

Little sun is an artist. Today he is playing mahjong alone. He suddenly feels that the tree in the yard doesn't look good. So he wants to decorate the tree.（The tree has n vertexs, indexed from 1 to n.）
Thought for a long time, finally he decides to use the mahjong to decorate the tree.
His mahjong is strange because all of the mahjong tiles had a distinct index.（Little sun has only n mahjong tiles, and the mahjong tiles indexed from 1 to n.）
He put the mahjong tiles on the vertexs of the tree.
As is known to all, little sun is an artist. So he want to decorate the tree as beautiful as possible.
His decoration rules are as follows:

(1)Place exact one mahjong tile on each vertex.
(2)The mahjong tiles' index must be continues which are placed on the son vertexs of a vertex.
(3)The mahjong tiles' index must be continues which are placed on the vertexs of any subtrees.

Now he want to know that he can obtain how many different beautiful mahjong tree using these rules, because of the answer can be very large, you need output the answer modulo 1e9 + 7.

 

Input

The first line of the input is a single integer T, indicates the number of test cases. 
For each test case, the first line contains an integers n. (1 <= n <= 100000)
And the next n - 1 lines, each line contains two integers ui and vi, which describes an edge of the tree, and vertex 1 is the root of the tree.

 

Output

For each test case, output one line. The output format is "Case #x: ans"(without quotes), x is the case number, starting from 1.

 

Sample Input

   
   
   
   

    
    
    
    2
9
2 1
3 1
4 3
5 3
6 2
7 4
8 7
9 3
8
2 1
3 1
4 3
5 1
6 4
7 5
8 4
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    Case #1: 32
Case #2: 16
   
   
   
   

 

Source

2015 Multi-University Training Contest 7

 

/* ***********************************************
Author        :CKboss
Created Time  :2015年08月12日 星期三 10时23分35秒
File Name     :HDOJ5379.cpp
************************************************ */

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>

#pragma comment(linker, "/STACK:1024000000,1024000000")

using namespace std;

typedef long long int LL;

const int maxn=100100;
const LL mod=1e9+7;

int n;

LL jc[maxn];
int sons[maxn],s[maxn];
LL dp[maxn];
vector<int> G[maxn];

void JC()
{
	jc[0]=1LL;
	for(LL i=1;i<maxn;i++) jc[i]=(jc[i-1]*i)%mod;
}

void init()
{
	for(int i=1;i<=n;i++) 
	{
		G[i].clear();
		sons[i]=0; s[i]=0; dp[i]=0;
	}
}

void check_dfs(int u,int fa)
{
	sons[u]=G[u].size();
	if(u!=fa) sons[u]--;
	if(sons[u]==0) return ;

	LL ts=0;
	for(int i=0,sz=G[u].size();i<sz;i++)
	{
		int v=G[u][i];
		if(v==fa) continue;
		check_dfs(v,u);
		if(sons[v]==0) ts++;
	}

	s[u]=sons[u]-ts;
}


void dfs(int u,int fa)
{
	if(sons[u]==0)
	{
		dp[u]=1LL;
		return ;
	}
	LL T=1;
	for(int i=0,sz=G[u].size();i<sz;i++)
	{
		int v=G[u][i];
		if(v==fa||v==0) continue;
		dfs(v,u);
		T=(T*dp[v])%mod;
	}
	if(s[u]==2)
	{
		dp[u]=T*jc[sons[u]-2]%mod;
	}
	else
	{
		dp[u]=2*T*jc[sons[u]-s[u]]%mod;
	}
}

int main()
{
	//freopen("in.txt","r",stdin);
	//freopen("out.txt","w",stdout);

	JC();
	int cas=1,T_T;
	scanf("%d",&T_T);
	while(T_T--)
	{
		scanf("%d",&n);
		init();
		for(int i=0,u,v;i<n-1;i++)
		{
			scanf("%d%d",&u,&v);
			G[u].push_back(v);
			G[v].push_back(u);
		}

		check_dfs(1,1);

		bool fg=true;
		for(int i=1;i<=n&&fg;i++)
			if(s[i]>2) fg=false;

		if(fg==false)
		{
			printf("Case #%d: 0\n",cas++);
			continue;
		}

		dfs(1,1);

		//for(int i=1;i<=n;i++) printf("%d: dp:%lld sons: %d s: %d\n",i,dp[i],sons[i],s[i]);

		printf("Case #%d: %d\n",cas++,(int)dp[1]);
	}
    
    return 0;
}