# Codeforces 855 C. Helga Hufflepuff’s Cup （树形dp）

## Description

Harry, Ron and Hermione have figured out that Helga Hufflepuff’s cup is a horcrux. Through her encounter with Bellatrix Lestrange, Hermione came to know that the cup is present in Bellatrix’s family vault in Gringott’s Wizarding Bank.

The Wizarding bank is in the form of a tree with total n vaults where each vault has some type, denoted by a number between 1 to m. A tree is an undirected connected graph with no cycles.

The vaults with the highest security are of type k, and all vaults of type k have the highest security.

There can be at most x vaults of highest security.

Also, if a vault is of the highest security, its adjacent vaults are guaranteed to not be of the highest security and their type is guaranteed to be less than k.

Harry wants to consider every possibility so that he can easily find the best path to reach Bellatrix’s vault. So, you have to tell him, given the tree structure of Gringotts, the number of possible ways of giving each vault a type such that the above conditions hold.

## Input

The first line of input contains two space separated integers, n and m — the number of vaults and the number of different vault types possible. (1 ≤ n ≤ 10^5, 1 ≤ m ≤ 10^9).

Each of the next n - 1 lines contain two space separated integers ui and vi (1 ≤ ui, vi ≤ n) representing the i-th edge, which shows there is a path between the two vaults ui and vi. It is guaranteed that the given graph is a tree.

The last line of input contains two integers k and x (1 ≤ k ≤ m, 1 ≤ x ≤ 10), the type of the highest security vault and the maximum possible number of vaults of highest security.

## Output

Output a single integer, the number of ways of giving each vault a type following the conditions modulo 10^9 + 7.

## Examples input

4 2
1 2
2 3
1 4
1 2


## Examples output

1


## 思路

$dp[i][j][k]$ 代表以 $i$ 为根的子树中共选择了 $j$ 个特殊颜色，且当前节点 $i$ 的状态为 $k$ 的染色方案数。

1. $k=0$ ，代表当前节点 $i$ 的颜色值小于 $K$ 。
2. $k=1$ ，代表当前节点 $i$ 的颜色值等于 $K$ 。
3. $k=2$ ，代表当前节点 $i$ 的颜色值大于 $K$ 。

for j in 0 -> sz[x]:

​ for k in 0 -> sz[to]:

$tmp[j+k][0]+=dp[x][j][0] \times (dp[to][k][0]+dp[to][k][1]+dp[to][k][2])%mod$

$tmp[j+k][1] += dp[x][j][1] \times dp[to][k][0]%mod$

$tmp[j+k][2] += dp[x][j][2] \times (dp[to][k][0]+dp[to][k][2])%mod$

## AC 代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5+10;
const int maxm = 20;
const int mod = 1e9+7;
typedef long long LL;

struct node
{
int to;
int next;
} edge[maxn<<1];

int n,m;
int K,X;
LL dp[maxn][maxm][3];
int sz[maxn];
int tmp[maxm][3];

void init()
{
memset(dp,0,sizeof(dp));
tot=0;
}

{
edge[tot].to=v;
}

void dfs(int x,int fa)
{
dp[x][0][0] = K-1;
dp[x][1][1] = 1;
dp[x][0][2] = m-K;
sz[x] = 1;
{
int to=edge[i].to;
if(to == fa)continue;
dfs(to,x);
memset(tmp,0,sizeof(tmp));
for(int j=0; j<=sz[x]; j++)
{
for(int k=0; k<=sz[to]; k++)
{
if(j+k>X) continue;
tmp[j+k][0] += dp[x][j][0]*(dp[to][k][0]+dp[to][k][1]+dp[to][k][2])%mod;
tmp[j+k][0]%=mod;
tmp[j+k][1] += dp[x][j][1]*dp[to][k][0]%mod;
tmp[j+k][1]%=mod;
tmp[j+k][2] += dp[x][j][2]*(dp[to][k][0]+dp[to][k][2])%mod;
tmp[j+k][2]%=mod;
}
}
sz[x] = min(sz[x]+sz[to],X);
for(int j=0; j<=sz[x]; j++)
for(int k=0; k<3; k++)
dp[x][j][k] = tmp[j][k];
}
}

int32_t main()
{
ios::sync_with_stdio(false);
cin.tie(NULL);
while(cin>>n>>m)
{
init();
for(int i=1; i<n; i++)
{
int u,v;
cin>>u>>v;