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.
.
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.
Example
1 4
2 1 3 2
1 2
1 3
3 4
8
0 3
1 2 3
1 2
2 3
3
4 8
7 8 7 5 4 6 4 10
1 6
1 2
5 8
1 3
3 5
6 7
3 4
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:求一棵树中满足特殊条件的连通子图的个数(该连通子图中节点最大值与最小值之差不超过d)
题解: Firstly, we solve the case d = + ∞. In this case, we can forget all ai since they doesn't play a role anymore. Consider the tree is rooted at node 1. Let Fi be the number of valid sets contain node i and several other nodes in subtree of i ("several" here means 0 or more). We can easily calculate Fi through Fj where j is directed child node of i: . Complexity: O(n).
General case: d ≥ 0. For each node i, we count the number of valid sets contain node i and some other node j such that ai ≤ aj ≤ ai + d (that means, node i have the smallest value a in the set). How? Start DFS from node i, visit only nodes j such that ai ≤ aj ≤ ai + d. Then all nodes have visited form another tree. Just apply case d = + ∞ for this new tree. We have to count n times, each time take O(n), so the overall complexity is O(n2). (Be craeful with duplicate counting)
意思是说,枚举最大值点(哇,又一次被打击了,满脑子骚操作,以为是DFS序,然后转区间巴拉巴拉的,结果直接枚举,orzzzzz)
并以该点为根跑一遍DFS,算满足条件的子树个数。区间重复问题:两个点a,b的值相同,当以a为根时,把b包含了,那么可能以b为
根时把a包含了,这样就可能重复一个区间,注意是可能。
#include<bits/stdc++.h> using namespace std; typedef long long ll; const int mod=1e9+7; int d,n; int a[2005],dp[2005]; bool vis[2005]; vector<int> G[2005]; void DFS(int u,int root){ dp[u]=1; vis[u]=true; for(int i=0;i<G[u].size();i++){ int v=G[u][i]; if(vis[v]) continue; if(a[v]==a[root]&&v<root) continue; //为了避免区间重复,规定方向,只能从小的编号走向大的。出题人就是强 if(a[v]<a[root]||a[v]>a[root]+d) continue; DFS(v,root); dp[u]=((ll)dp[u]*(dp[v]+1))%mod; } } int main() { cin>>d>>n; for(int i=1;i<=n;i++) cin>>a[i]; for(int i=2;i<=n;i++){ int p,q; cin>>p>>q; G[p].push_back(q); G[q].push_back(p); } int ans=0; for(int i=1;i<=n;i++){ for(int j=1;j<=n;j++){ dp[j]=0; vis[j]=false; } DFS(i,i); ans=(ans+dp[i])%mod; } cout<<ans<<endl; }