CodeForces 161D Distance in Tree

Distance in Tree

Time Limit: 3000ms

Memory Limit: 524288KB

This problem will be judged on CodeForces. Original ID: 161D
64-bit integer IO format: %I64d      Java class name: (Any)

 

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 %I64d specifier.

 

Sample Input

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

Hint

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

 

Source

VK Cup 2012 Round 1

 

解题：树形dp dp[u][i]表示从子树到u为根长度为i的路径数目

 

#include <bits/stdc++.h>
using namespace std;
const int maxn = 50010;
vector<int>g[maxn];
int dp[maxn][505],n,k,ret;
void dfs(int u,int fa){
    for(int i = 1; i <= k; ++i) dp[u][i] = 0;
    dp[u][0] = 1;
    for(int i = g[u].size()-1; i >= 0; --i){
        if(g[u][i] == fa) continue;
        dfs(g[u][i],u);
        for(int j = 0; j < k; ++j) ret += dp[u][j]*dp[g[u][i]][k - j - 1];
        for(int j = 1; j <= k; ++j) dp[u][j] += dp[g[u][i]][j-1];
    }
}
int main(){
    int u,v;
    while(~scanf("%d%d",&n,&k)){
        for(int i = ret = 0; i < maxn; ++i) g[i].clear();
        for(int i = 1; i < n; ++i){
            scanf("%d%d",&u,&v);
            g[u].push_back(v);
            g[v].push_back(u);
        }
        dfs(1,0);
        printf("%d
",ret);
    }
    return 0;
}