codeforces 161D. Distance in Tree（树dp）

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#树dp #乘法原理

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本文介绍了一种使用树形动态规划（树DP）的方法来解决在一个给定树中寻找具有特定距离k的顶点对数量的问题。通过定义状态dp[u][j]为以u为根的子树中到u的距离为j的节点数量，并利用递归思想更新这些值，最终能够高效地计算出所需答案。

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.

   Sample test(s)
 

     input
   
5 2
1 2
2 3
3 4
2 5

     output
   
4

     input
   
5 3
1 2
2 3
3 4
4 5

     output
   
2

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：给出一棵树，每条边的长度都为1。现在给出一个k，问树中距离为k的两个点共有多少对。

思路：树dp。设dp[u][j]为以u为根的子树中到u的距离为j的结点个数，运用乘法原理更新dp值。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <cmath>
#define ll long long
#define L(rt) (rt<<1)
#define R(rt) (rt<<1|1)
using namespace std;

const int maxn = 50005;
const int maxm = 1000005;
const int INF = 1 << 20;
int n, k, ans;
int dp[maxn][505];
vector<int> G[maxn];
void dfs(int u, int pre){
    memset(dp[u], 0, sizeof(dp[u]));
    dp[u][0] = 1;
    for(int i = 0; i < (int)G[u].size(); i++)
    {
        int v = G[u][i];
        if(v == pre) continue;
        dfs(v, u);
        for(int j = 1; j <= k; j++)
        ans += dp[u][j - 1] * dp[v][k - j];
        for(int j = 1; j <= k; j++)
        dp[u][j] += dp[v][j - 1];
    }
}
int main()
{
    int a, b;
    while(~scanf("%d%d", &n, &k))
    {
        for(int i = 0; i <= n; i++) G[i].clear();
        for(int i = 0; i < n - 1; i++)
        {
            scanf("%d%d", &a, &b);
            G[a].push_back(b);
            G[b].push_back(a);
        }
        ans = 0;
        dfs(1, -1);
        printf("%d\n", ans);
    }
    return 0;
}