HDU 5379(Mahjong tree-树形dp统计标号)
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.
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
Author
UESTC
Source
2015 Multi-University Training Contest 7
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显然，如果有大于2个非叶子节点肯定无解，
l1叶子节点可以全排列l1!
如果有一个非叶子节点，可能在两端，答案*2
如果有两个非叶子节点，必然各在一端，答案*2
最后由于当前根节点要么在首要么在尾答案*2，但由于这2种情况在父节点的时候才被决定，所以不用管
#include<bits/stdc++.h>
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<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;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\n",kcase); continue;
}
printf("Case #%d: %I64d\n",kcase,mul(dfs(1,0),2));
}
return 0;
}