HDU 5379 (Mahjong tree-tree dp statistical label)

HDU 5379 (Mahjong tree-tree dp statistical label)

 

Mahjong tree Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission (s): 1460 Accepted Submission (s): 445


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

Author UESTC
Source 2015 Multi-University Training Contest 7
Recommend wange2014 | We have carefully selected several similar problems for you: 5421 5420 5419 5418

 

 

Obviously, if there are more than two non-leaf nodes,

L1 leaf nodes can all be arranged in l1!

If there is a non-leaf node, it may be at both ends. Answer * 2

If there are two non-leaf nodes, each of them must be at one end. Answer: * 2

At last, because the current root node is either at the beginning or at the end of the answer * 2, but the two cases are determined at the parent node, you do not need to worry about

 

 

 

 

 

#include
 
   using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define Rep(i,n) for(int i=0;i
  
   =0;i--)#define Forp(x) for(int p=pre[x];p;p=Next[p])#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])  #define Lson (x<<1)#define Rson ((x<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,127,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define INF (2139062143)#define F (1000000007)#define MAXN (200000+10)#define MAXM (200000+10)typedef long long ll;ll mul(ll a,ll b){return (a*b)%F;}ll add(ll a,ll b){return (a+b)%F;}ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}void upd(ll &a,ll b){a=(a%F+b%F)%F;}int n;int pre[MAXN],Next[MAXM],edge[MAXM],siz=1;void addedge(int u,int v) {edge[++siz]=v;Next[siz]=pre[u];pre[u]=siz;} void addedge2(int u,int v){addedge(u,v),addedge(v,u);}int sz[MAXN];ll jie[MAXN];void init(){jie[0]=1;For(i,200000) jie[i]=(jie[i-1]*i)%F;} ll dfs(int x,int fa){int l1=0,l2=0;ll ans=1;Forp(x) {int v=edge[p];if (v==fa) continue;ans=mul(ans,dfs(v,x));sz[x]+=sz[v];if (sz[v]==1) l1++;else l2++;}sz[x]++;if (l2>2) {return 0;} if (l1+l2==0) return 1;if (l2==0) return ans=mul(ans,mul(jie[l1],1));if (l2==1) return ans=mul(ans,mul(jie[l1],2));if (l2==2) return ans=mul(ans,mul(jie[l1],2));}int main(){//freopen(K.in,r,stdin);init(); int T;cin>>T;For(kcase,T) {MEM(pre) MEM(Next) MEM(edge) siz=1;MEM(sz)scanf(%d,&n);For(i,n-1) {int a,b;scanf(%d%d,&a,&b);addedge2(a,b);}if (n==1) {printf(Case #%d: 1,kcase); continue;}printf(Case #%d: %I64d,kcase,mul(dfs(1,0),2));}return 0;}