hdu5379Mahjong tree

题意：给出一棵树，用[1,n]里的n个不同的数去给结点标号，要求任意结点的儿子标号都是连续的，任意子树里的标号也都是连续的。求这样标号的方法有多少种。

做法：很显然，若一个结点有大于两个的儿子必然无解。

若当前有x个连续的数可以用来分配，那么这个结点要么取最小的那个数，要么取最大的那个数，

当该结点刚好有两个不为叶子的儿子，然后剩下的那x-1个数要么把前面的部分给左边的儿子要么给右边的儿子，后面的部分一样，这就有2种可能。

当刚好有一个不为叶子的儿子，那么剩下的那x-1个数要么把前面的部分给儿子，要么把后面的部分给儿子，这就有2种可能。

若没有上述的儿子，则全是叶子，则就是(x-1)!可能，上述两种情况考虑儿子为叶子的时候同理。

接下来就是很裸的树形dp了。

#pragma comment(linker, "/STACK:102400000,102400000")#include<iostream>#include<cstdio>#include<cstring>#include<vector>using namespace std;typedef long long ll;const ll mod=1000000007;struct Edge{    int to,next;}edge[200010];int head[100010],tail;void add(int from,int to){    edge[tail].to=to;    edge[tail].next=head[from];    head[from]=tail++;}bool vis[100010];int num[100010];void seek(int from){    vis[from]=1;    num[from]=1;    for(int i=head[from];i!=-1;i=edge[i].next)    {        int to=edge[i].to;        if(vis[to])            continue;        seek(to);        num[from]+=num[to];    }}ll fac[100010];ll dfs(int from){    vis[from]=1;    int sum1=0,sum2=0;    vector<int>bx;    for(int i=head[from];i!=-1;i=edge[i].next)    {        int to=edge[i].to;        if(vis[to])            continue;        if(num[to]==1)            sum1++;        else        {            sum2++;            bx.push_back(to);        }    }    if(sum2>2)    {        return 0;    }    if(sum2==0)        return fac[sum1];    ll sum=1;    for(int i=0;i<sum2;i++)    {        ll t=dfs(bx[i]);        if(t==0)            return 0;        sum=(sum*t)%mod;    }    return fac[sum1]*sum*2%mod;}int main(){    fac[0]=1;    for(int i=1;i<=100000;i++)        fac[i]=fac[i-1]*i%mod;    int T;    scanf("%d",&T);    for(int cs=1;cs<=T;cs++)    {        int n;        scanf("%d",&n);        tail=0;        memset(head,-1,sizeof(head));        for(int i=1;i<n;i++)        {            int u,v;            scanf("%d%d",&u,&v);            add(u,v);            add(v,u);        }        if(n==1)        {            printf("Case #%d: 1\n",cs);            continue;        }        memset(vis,0,sizeof(vis));        seek(1);        memset(vis,0,sizeof(vis));        printf("Case #%d: %d\n",cs,int(2*dfs(1)%mod));    }    return 0;}

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 999    Accepted Submission(s): 318

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

292 13 14 35 36 27 48 79 382 13 14 35 16 47 58 4

 

Sample Output

Case #1: 32Case #2: 16

 

Source

2015 Multi-University Training Contest 7