题目链接： pid=5379">http://acm.hdu.edu.cn/showproblem.php? pid=5379
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.
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.
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
题意：
一颗树共同拥有 n 个节点，现要把他们从 1 - n 编号。
条件：
1、每一个节点的子节点之间号是连续的（如：4 5 6 或者6 5 4均可）！
2、每棵子树的编号自身是连续的！
求一共同拥有多少种可能的编号方式。
PS:
假设一个节点的子树中规模大于1的子树多于2个。那么肯定是不可能有满足要求的安排编号的！
在一个区间中。规模大于1的子树必定选择的是最左边或者最右边的连续的编号。
剩下规模为1的子树，有N!种安排方法。
手动扩栈，须要用C++提交！
代码例如以下：
#pragma comment(linker, "/STACK:102400000,102400000")
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;
#define LL long long
#define maxn 100017
const LL mod = 1e9+7;
vector<int> v[maxn];
LL f[maxn];
LL son[maxn];
void init()
{
f[0] = 1;
for(int i = 1; i <= maxn; i++)
{
f[i] = (i*f[i-1])%mod;
}
}
LL dfs(int u, int father)
{
son[u] = 1;
int x = 0, y = 0;
LL ans = 1;
int num = v[u].size();
for(int i = 0; i < num; i++)
{
if(v[u][i] == father)
{
continue;
}
int t1 = v[u][i];
LL tt = dfs(t1, u);
ans = (ans*tt)%mod;
if(son[v[u][i]] >= 2)//规模大于1的子树
{
x++;
}
else
{
y++;
}
if(ans==0 || x > 2)//ans==0无解，规模大于1的子树多于2个肯定不可能
{
return 0;
}
son[u]+=son[v[u][i]];
}
if(x)
{
ans*=2;
ans%=mod;
}
ans = (ans*f[y])%mod;
return ans;
} int main()
{
int t;
int n;
int cas = 0;
init();
scanf("%d",&t);
while(t--)
{
memset(son,0,sizeof(son));
scanf("%d",&n);
for(int i = 0; i < maxn; i++)
{
v[i].clear();
}
int x, y;
for(int i = 1; i < n; i++)
{
scanf("%d%d",&x,&y);
v[x].push_back(y);
v[y].push_back(x);
}
LL ans = dfs(1, -1);
if(son[1] > 1)
{
ans*=2;
ans%=mod;
}
printf("Case #%d: %I64d\n",++cas,ans);
}
return 0;
}