Manthan, Codefest 17: C. Helga Hufflepuff's Cup（树形DP）

C. Helga Hufflepuff's Cup

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 ≤ 105, 1 ≤ m ≤ 109).

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 109 + 7.

Examples

input

4 21 22 31 41 2

output

1

input

3 31 21 32 1

output

13

input

3 11 21 31 1

output

0

题意：

n个节点的树，你要为树上的每一个节点添加一个权值（范围为[1, m]），要求其中权值为k的节点不超过x个，并且如果一个节点权值为k，周围所有的节点权值一定都<k，问有多少种合法方案

dp[u][x][0]表示以第u个节点为根的子树中，权值为k的节点刚好有x个，并且u点权值<k的方案数

dp[u][x][1]表示以第u个节点为根的子树中，权值为k的节点刚好有x个，并且u点权值=k的方案数

dp[u][x][2]表示以第u个节点为根的子树中，权值为k的节点刚好有x个，并且u点权值>k的方案数

dp时归并所有子树

#include<stdio.h>#include<vector>#include<string.h>using namespace std;vector<int> G[100005];#define mod 1000000007#define LL long longLL m, k, p, dp[100005][12][3], size[100005], temp[12][3];void Sech(LL u, LL fa){LL i, j, x, v;dp[u][0][0] = k-1;dp[u][1][1] = 1;dp[u][0][2] = m-k;size[u] = 1;for(i=0;i<G[u].size();i++){v = G[u][i];if(v==fa)continue;Sech(v, u);memset(temp, 0, sizeof(temp));for(j=0;j<=size[u];j++){for(x=0;x<=size[v];x++){if(j+x>p)continue;temp[j+x][0] = (temp[j+x][0]+dp[u][j][0]*(dp[v][x][0]+dp[v][x][1]+dp[v][x][2])%mod)%mod;temp[j+x][1] = (temp[j+x][1]+dp[u][j][1]*dp[v][x][0])%mod;temp[j+x][2] = (temp[j+x][2]+dp[u][j][2]*(dp[v][x][0]+dp[v][x][2])%mod)%mod;}}size[u] = min(size[u]+size[v], p);for(j=0;j<=size[u];j++){dp[u][j][0] = temp[j][0];dp[u][j][1] = temp[j][1];dp[u][j][2] = temp[j][2];}}}int main(void){LL n, i, x, y, ans;scanf("%I64d%I64d", &n, &m);for(i=1;i<=n-1;i++){scanf("%I64d%I64d", &x, &y);G[x].push_back(y);G[y].push_back(x);}scanf("%I64d%I64d", &k, &p);Sech(1, -1);ans = 0;for(i=0;i<=p;i++)ans = (ans+dp[1][i][0]+dp[1][i][1]+dp[1][i][2])%mod;printf("%I64d\n", ans);return 0;}