D. Distance in Tree

最新推荐文章于 2022-03-30 20:26:07 发布

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最新推荐文章于 2022-03-30 20:26:07 发布

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分类专栏： cf dp

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给定一棵树和一个正整数k，求树中距离为k的节点对的数量。题目要求输出不同节点对的数量，注意排列与组合的区别。可以通过树形动态规划解决此问题。

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A tree is a connected graph that doesn't contain any cycles.

The distance between two vertices of a tree is the length (in edges) of the shortest path between these vertices.

You are given a tree with n vertices and a positive number k. Find the number of distinct pairs of the vertices which have a distance of exactly k between them. Note that pairs (v, u) and (u, v) are considered to be the same pair.

Input

The first line contains two integers n and k (1 ≤ n ≤ 50000, 1 ≤ k ≤ 500) — the number of vertices and the required distance between the vertices.

Next n - 1 lines describe the edges as "ai bi" (without the quotes) (1 ≤ ai, bi ≤ n, ai ≠ bi), where ai and bi are the vertices connected by the i-th edge. All given edges are different.

Output

Print a single integer — the number of distinct pairs of the tree's vertices which have a distance of exactly k between them.

Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64dspecifier.

Examples

input

Copy

output

Copy

input

Copy

output

Copy

Note

In the first sample the pairs of vertexes at distance 2 from each other are (1, 3), (1, 5), (3, 5) and (2, 4).

题意：

给你一棵树，要你求顶点之间距离为k的有多少对，求的是组合，不是排列。

思路：

树型dp，设计状态，为dp[u][j],以u为起点的长度为j的顶点个数。

代码：

#include<bits/stdc++.h>
#define LL long long

using namespace std;
const int maxn=5e4+100;
const int maxm=600;
LL ans;
struct node
{
    int to,Next;
};
node Edge[2*maxn];
int Head[maxn];
int cnt;
int dp[maxn][maxm];
void add(int u,int v)
{
    Edge[++cnt].to=v;
    Edge[cnt].Next=Head[u];
    Head[u]=cnt;
    return ;
}
void dfs(int u,int fa,int k)
{
    dp[u][0]=1;
    for(int i=1;i<=k;i++)
    {
        dp[u][i]=0;
    }
    for(int i=Head[u];i!=-1;i=Edge[i].Next)
    {
        int v=Edge[i].to;
        if(v==fa)
        {
            continue;
        }
        dfs(v,u,k);
        for(int j=0;j<k;j++)
        {
            ans+=dp[u][j]*dp[v][k-j-1];
        }
        for(int j=1;j<=k;j++)
        {
            dp[u][j]+=dp[v][j-1];
        }
    }
    return;
}
int main()
{
    memset(Head,-1,sizeof(Head));
    int n,k;
    cin>>n>>k;
    for(int i=1;i<n;i++)
    {
        int u,v;
        cin>>u>>v;
        add(u,v);
        add(v,u);
    }
    dfs(1,0,k);
    cout<<ans<<endl;
    return 0;
}

反思：这种树上计数的题总能想到大概的思路，但总是写不出要好好练习一下。