Mahjong tree (hdu 5379 dfs)

Mahjong tree

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 190    Accepted Submission(s): 58

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

 

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题意：一颗n个节点n-1条边的树，如今要给每一个节点标号（1~n），要求：（1）每一层的兄弟节点的标号要是连续的（2）每一颗子树的全部节点标号是连续的。问有多少种标号方案。

思路：对于每一层顶多仅仅能存在2个非叶子节点。否则无解；对于每一层有x个叶子节点，y个非叶子节点，那么ans=(ans * x!)%mod，另外假设y!=0，还得ans=2*ans%mod。

代码：

#include <iostream>
#include <functional>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define FRE(i,a,b)  for(i = a; i <= b; i++)
#define FREE(i,a,b) for(i = a; i >= b; i--)
#define FRL(i,a,b)  for(i = a; i < b; i++)
#define FRLL(i,a,b) for(i = a; i > b; i--)
#define mem(t, v)   memset ((t) , v, sizeof(t))
#define sf(n)       scanf("%d", &n)
#define sff(a,b)    scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf          printf
#define DBG         pf("Hi\n")
typedef long long ll;
using namespace std;

#define INF 0x3f3f3f3f
#define mod 1000000007
const int maxn = 1005;
const int MAXN = 100001;
const int MAXM = 200002;
const int N = 1005;

struct Edge
{
    int u,v,next;
}edge[MAXM];

int tot,head[MAXN],n;
ll fn[MAXN],ans;
bool flag;

void INIT()
{
    fn[0]=1;
    for (int i=1;i<=100000;i++)
        fn[i]=(fn[i-1]*(ll)i)%mod;
}

void init()
{
    tot=0;flag=true;ans=1;
    memset(head,-1,sizeof(head));
}

void addedge(int u,int v)
{
    edge[tot].u=u;
    edge[tot].v=v;
    edge[tot].next=head[u];
    head[u]=tot++;
}

ll dfs(int u,int pre)
{
    if (!flag) return 0;
    ll s=1,all=0,ss=0;
    for (int i=head[u];~i;i=edge[i].next)
    {
        int v=edge[i].v;
        if (v==pre) continue;
        ll x=dfs(v,u);
        s+=x;
        if (x==1) all++;
        else if (x>1) ss++;
    }
    if (ss>2) flag=false;
    else{
        ans=(ans*fn[all])%mod;
        if (ss!=0) ans=((ll)2*ans)%mod;
    }
    return s;
}

int main()
{
#ifndef ONLINE_JUDGE
    freopen("C:/Users/lyf/Desktop/IN.txt","r",stdin);
#endif
    int i,j,t,u,v,cas=0;
    INIT();
    scanf("%d",&t);
    while (t--)
    {
        scanf("%d",&n);
        init();
        for (i=0;i<n-1;i++)
        {
            scanf("%d%d",&u,&v);
            addedge(u,v);
            addedge(v,u);
        }
        dfs(1,-1);
        if (n>1) ans=((ll)2*ans)%mod;
        if (!flag) ans=0;
        printf("Case #%d: %lld\n",++cas,ans);
    }
    return 0;
}