codeforces contest 855 problem C(树形DP)

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 2
1 2
2 3
1 4
1 2

Output

1

Input

3 3
1 2
1 3
2 1

Output

13

Input

3 11 2
1 3
1 1

Output

0

Note
In test case 1, we cannot have any vault of the highest security as its type is
1 implying that its adjacent vaults would have to have a vault type less than
1, which is not allowed. Thus, there is only one possible combination, in which all the vaults have type
2.

这题主要是 DP数组及状态不好想，构造出DP数组 状态转移方程比较容易理解

每一个节点都可以有m种取法 对他有约束作用的就是相邻子树 以u为根的节点能有多少种取法 取决于子树有多少种取法 类似于背包 将最大的限制数看成是一个空间为x的背包 每次枚举子节点 更新背包的方案数

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 <algorithm>
#include <string.h>
#include <iostream>
#include <stdio.h>
#include <string>
#include <vector>
#include <queue>
#include <map>
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N = 1e5+10;
const LL mod = 1e9+7;
LL dp
[11][4],tmp[11][4],size1;
vector<int>p
;
LL n, m, k, x;
void dfs(int u,int pre)
{
dp[u][0][0]=k-1,dp[u][1][1]=1,dp[u][0][2]=m-k;
size1[u]=1;
for(int i=0;i<p[u].size();i++)
{
int v=p[u][i];
if(v==pre) continue;
dfs(v,u);
memset(tmp,0,sizeof(tmp));
for(int j=0;j<=size1[u];j++)
{
for(int h=0;h<=size1[v];h++)
{
if(j+h>x)break;
tmp[j+h][0]=(tmp[j+h][0]+dp[u][j][0]*(dp[v][h][0]+dp[v][h][1]+dp[v][h][2]))%mod;
tmp[j+h][1]=(tmp[j+h][1]+dp[u][j][1]*(dp[v][h][0]))%mod;
tmp[j+h][2]=(tmp[j+h][2]+dp[u][j][2]*(dp[v][h][0]+dp[v][h][2]))%mod;
}
}
size1[u]=min(size1[u]+size1[v],x);
for(int j=0;j<=size1[u];j++)
dp[u][j][0]=tmp[j][0],dp[u][j][1]=tmp[j][1],dp[u][j][2]=tmp[j][2];
}
return ;
}

int main()
{
scanf("%I64d %I64d",&n,&m);
for(int i=0; i<=n; i++)p[i].clear();
for(int i=1; i<n; i++)
{
int x,y;
scanf("%d %d", &x, &y);
p[x].push_back(y),p[y].push_back(x);
}
scanf("%I64d %I64d",&k,&x);
dfs(1,-1);
LL ans=0;
for(int i=0;i<=x;i++) ans=(ans+dp[1][i][0]+dp[1][i][1]+dp[1][i][2])%mod;
cout<<ans<<endl;
return 0;
}