codeforces 161D. Distance in Tree（树dp）

D. Distance in Tree

time limit per test
3 seconds

memory limit per test
512 megabytes

input
standard input

output
standard output

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 22 3
3 44 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;
}