codeforces 486d Valid Sets

最新推荐文章于 2020-07-29 21:00:47 发布

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最新推荐文章于 2020-07-29 21:00:47 发布

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本文介绍了一种算法，用于解决给定一棵树后，寻找所有满足特定条件的连通子图的问题。具体条件为子图中最大节点值与最小节点值之差不大于给定值d。文章详细阐述了解决方案的思路，并提供了实现该算法的C++代码。

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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.
$\text{[math]}$ .

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.

Examples
inputCopy
1 4
2 1 3 2
1 2
1 3
3 4
output
8
inputCopy
0 3
1 2 3
1 2
2 3
output
3
inputCopy
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：给出一棵树，求这棵树的满足最大点与最小点之差小于d的连通子图的个数。

题解：我们首先选择一个点，把这个点当做最大值以及根，然后往下考虑。

对于一个点，要么选要么不选，选了的话就递归，不选的话他的子树的点都不选。

有可能因为选择相等的点而重复，所以我们每次都dfs权值相等的点时，只考虑编号更大的点。

代码：

#include <cstdio>
#include <vector>
#include <cstring>
using namespace std;
long long d,dp[2005],V[2005],n,u,v,ans;
vector<int>G[2005];
void dfs(int s,int fa,int rt)
{
    dp[s]=1;
    for(int i=0,to;i<G[s].size();i++)
    {
        to=G[s][i];
        if(to==fa) continue;
        dfs(to,s,rt);                          
        if((V[rt]>V[to]&&V[rt]-V[to]<=d)||(V[rt]==V[to]&&rt<to)) dp[s]=(dp[s]+dp[s]*dp[to])%1000000007; 
    }                                          
}
int main()
{
    scanf("%I64d%I64d",&d,&n);
    for(int i=1;i<=n;i++) scanf("%I64d",&V[i]);
    for(int i=1;i<n;i++)
    {
        scanf("%I64d%I64d",&u,&v);
        G[u].push_back(v);
        G[v].push_back(u);
    }
    for(int i=1;i<=n;i++)
    {                     
        memset(dp,0,sizeof(dp));
        dfs(i,-1,i);
        ans=(ans+dp[i])%1000000007;
    }
    printf("%I64d\n",ans);
}