HDU 5379 Mahjong tree

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

理解题意，一个节点的儿子节点中还有儿子节点的点不会超过2个，否则就无解，然后分类讨论一下，一次dfs可以解决

#pragma comment(linker, "/STACK:102400000,102400000")
#include<cstdio>
#include<cmath>
#include<queue>
#include<stack>
#include<string>
#include<vector>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long LL;
const int maxn=100005;
const LL base=1e9+7;
vector<int> tree[maxn];
int T,n,x,y,tt=0;
LL ans[maxn],f[maxn],flag;

void dfs(int x,int fa)
{
    if (!flag) return ;
    LL a=tree[x].size(),b=0;
    ans[x]=1;
    for (int i=0;i<a;i++)
    if (tree[x][i]!=fa)
    {
        dfs(tree[x][i],x);
        if (!flag) return ;
        ans[x]=(ans[x]*ans[tree[x][i]])%base;
        b+=(tree[tree[x][i]].size()>1)?1:0;
    }
    if (b>2) {flag=0; return ;}
    ans[x]=ans[x]*f[a-1-b]%base;
    if (b) ans[x]=(ans[x]<<1)%base;
}

int main()
{
    for (LL i=f[0]=1;i<=100000;i++) f[i]=f[i-1]*i%base;
    scanf("%d",&T);
    while (T--)
    {
        flag=1;
        scanf("%d",&n);
        if (n==1){ printf("Case #%d: 1\n",++tt); continue;}
        for (int i=1;i<=n;i++) tree[i].clear();
        tree[1].push_back(1);
        for (int i=1;i<n;i++)
        {
            scanf("%d%d",&x,&y);
            tree[x].push_back(y);
            tree[y].push_back(x);
        }
        dfs(1,1);
        printf("Case #%d: %lld\n",++tt,flag?((ans[1]<<1)%base):flag);
    }
    return 0;
}