hdu5379 Mahjong tree DFS 多校联合第七场

Mahjong tree

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 939    Accepted Submission(s): 294

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
   
   
   
   

  一棵树N个结点，编号1-N，问有多少种编号方法满足：每个节点的子结点编号是连续的，任意一棵子树的编号是连续的。

  以某个结点为根的子树，假设根节点编号已经确定，这个根节点最多只能有两个非叶子儿子，因为非叶子儿子编号取值必须是整个编号范围的两端，所以如果有结点的非叶子儿子大于两个的话肯定是不行的。刚开始做的时候dfs还加上了编号范围，但这是没必要的，因为只要是一段连续的数，种数都是相同的。用dfs(u)表示以u为根的树有多少种满足的编号情况（u的编号已经确定，并且剩下结点编号是连续的一段）。非叶子的儿子只能取两端的编号，接着dfs，叶子编号是可以随意排的，所以再乘上叶子个数的阶乘。注意只有一个结点的情况，特判一下。

#include<cstring>
#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<queue>
#define INF 0x3f3f3f3f
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;

typedef long long LL;

const int MAXN=100010;
const LL MOD= 1000000007;

int T,N;
LL fac[MAXN];
int cnt[MAXN],num[MAXN];
vector<int> g[MAXN];

int dfs1(int u,int fa){
    int len=g[u].size();
    num[u]=1;
    cnt[u]=0;
    for(int i=0;i<len;i++){
        int v=g[u][i];
        if(v!=fa){
            int n=dfs1(v,u);
            num[u]+=n;
            if(n>1) cnt[u]++;
        }
    }
    return num[u];
}

LL dfs2(int u,int fa){
    int len=g[u].size();
    if(cnt[u]>2) return 0;
    int lef=0,v1=-1,v2=-1;
    for(int i=0;i<len;i++){
        int v=g[u][i];
        if(v!=fa){
            if(num[v]<=1) lef++;
            else{
                if(v1==-1) v1=v;
                else v2=v;
            }
        }
    }
    LL ret=0;
    LL m=fac[lef];
    if(v1!=-1){
        if(v2!=-1) ret=(ret+m*dfs2(v1,u)%MOD*dfs2(v2,u)%MOD*2%MOD)%MOD;
        else ret=(ret+m*dfs2(v1,u)%MOD*2%MOD)%MOD;
    }
    else ret=m;
    return ret;
}

int main(){
    freopen("in.txt","r",stdin);
    scanf("%d",&T);
    int cas=0;
    fac[0]=1;
    for(int i=1;i<100010;i++) fac[i]=fac[i-1]*i%MOD;
    while(T--){
        scanf("%d",&N);
        int u,v;
        for(int i=1;i<=N;i++) g[i].clear();
        for(int i=0;i<N-1;i++){
            scanf("%d%d",&u,&v);
            g[u].push_back(v);
            g[v].push_back(u);
        }
        memset(cnt,0,sizeof(cnt));
        dfs1(1,-1);
        printf("Case #%d: ",++cas);
        if(N==1) printf("1\n");
        else printf("%I64d\n",2*dfs2(1,-1)%MOD);
    }
    return 0;
}