Codeforces Round #277 (Div. 2)D（树形DP计数类）树形DP

D. Valid Sets time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output As you know, an undirected connected graph with n nodes and n?-?1 edges is called a tree. You are given an integer d and a tree consisting of n nodes. Each node i has a value ai associated with it.
We call a set S of tree nodes valid if following conditions are satisfied:

S is non-empty.
S is connected. In other words, if nodes u and v are in S, then all nodes lying on the simple path between u and v should also be presented in S.
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【Codeforces Round #277 (Div. 2)D（树形DP计数类）】 Your task is to count the number of valid sets. Since the result can be very large, you must print its remainder modulo 1000000007(109?+?7).
Input The first line contains two space-separated integers d (0?≤?d?≤?2000) and n (1?≤?n?≤?2000).
The second line contains n space-separated positive integers a1,?a2,?...,?an(1?≤?ai?≤?2000).
Then the next n?-?1 line each contain pair of integers u and v (1?≤?u,?v?≤?n) denoting that there is an edge between u and v. It is guaranteed that these edges form a tree.
Output Print the number of valid sets modulo 1000000007.
Sample test(s) input

1 4 2 1 3 2 1 2 1 3 3 4

output

8

input

0 3 1 2 3 1 2 2 3

output

3

input

4 8 7 8 7 5 4 6 4 10 1 6 1 2 5 8 1 3 3 5 6 7 3 4

output

41

Note In the first sample, there are exactly 8 valid sets: {1},?{2},?{3},?{4},?{1,?2},?{1,?3},?{3,?4} and {1,?3,?4}. Set {1,?2,?3,?4} is not valid, because the third condition isn't satisfied. Set {1,?4} satisfies the third condition, but conflicts with the second condition.

题意：RT
思路：树形DP，dp[u]表示以u为根，且它的所有子树的点v的权值都不超过它，且满足 w[u]-w[v]<=d 的方案数
对于每个点为根这样算一遍就好了，注意减去算重的情况，即相邻多个点的权值相等，可以用点的代号去重（即规定只能从代号大的往小的DP）

#include #include #include #include #include #include #include #include using namespace std; typedef long long ll; const int MAXN = 2005; const int MOD = (int)1e9+7; int edge[MAXN<<1]; int next[MAXN<<1]; int head[MAXN]; int esz; int ww[MAXN]; ll dp[MAXN]; int D; void addedge(int u,int v) { edge[esz]=v; next[esz]=head[u]; head[u]=esz++; }void dfs(int u,int p,int w,int root) { if((w>ww[u]&&w-ww[u]<=D) || (w==ww[u]&&root>=u)){dp[u]=1; for(int i=head[u]; i!=-1; i=next[i]){ int v=edge[i]; if(v==p)continue; dfs(v,u,w,root); dp[u]=dp[u]*((dp[v]+1)%MOD)%MOD; } } }int main() { int n; scanf("%d%d",&D,&n); memset(head,-1,sizeof(head)); esz=0; for(int i=1; i<=n; i++) scanf("%d",&ww[i]); for(int i=1; i