[codeforces 161D] Distance in Tree（树的点分治）

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给定一棵树和一个正整数k，要求找出树中距离为k的节点对的数量。输入包含树的节点数和k值，以及描述树边的连接信息。输出为符合条件的节点对数。对于给定的例子，当k=2时，第一棵树中有4对节点距离为2，第二棵树中有2对节点距离为3。

题目：http://codeforces.com/problemset/problem/161/D

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 %I64d specifier.
Examples
Input

5 2
1 2
2 3
3 4
2 5

Output

Input

5 3
1 2
2 3
3 4
4 5

Output

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：
给一棵树，统计距离为k的点对数。

思路：和上一题一样，只不过统计是＝ｋ的对数；
在ｃａｌ函数中引入 tmp[],cnt[]，代表作到ｘ子树时，当前层ｓｏｌｖｅ中之前子树《＝ｋ的ｄｉｓ个数，因为k<=500;Ｏ（ｎ）计算，且不重复；

代码：

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<vector>
#define N 50005
using namespace std;
int n,K,rt,sz[N];
int cnt[505],tmp[505];
vector<int> lin[N];
int vis[N],f[N];
int size;
long long  ans;
int getrt(int x,int fa)
{
    sz[x]=1;
    f[x]=0;
    for(int i=0;i<lin[x].size();i++)
    {
        int u=lin[x][i];
        if(u==fa||vis[u]) continue;
        getrt(u,x);
        sz[x]+=sz[u];
        f[x]=max(f[x],sz[u]);
    }
    f[x]=max(f[x],size-f[x]);
    if(f[rt]>f[x]) rt=x;

}
void getdis(int x,int fa,int dis)
{
    sz[x]=1;
    if(dis<=K)
    {   
        ans+=cnt[K-dis];
        ++tmp[dis];
    }
    for(int i=0;i<lin[x].size();i++)
    {
        int u=lin[x][i];
        if(vis[u]||u==fa) continue;
        getdis(u,x,dis+1);
        sz[x]+=sz[u];
    }
}
void cal(int x)
{
    memset(cnt,0,sizeof(cnt));
    int ret=0;
    cnt[0]=1;
    for(int i=0;i<lin[x].size();i++)
    {
        int u=lin[x][i];
        if(vis[u]) continue;
        memset(tmp,0,sizeof(tmp));
        getdis(u,u,1);
        for(int j=1;j<=K;j++)
        cnt[j]+=tmp[j];

    }
}
void solve(int x)
{
    vis[x]=1;
    cal(x);
    for(int i=0;i<lin[x].size();i++)
    {
        int u=lin[x][i];
        if(vis[u]) continue;    
        f[0]=size=sz[u];
        getrt(u,rt=0);
        solve(rt);
    }
}
int main()
{
    int aa,bb;
    scanf("%d%d",&n,&K);
    for(int i=1;i<n;i++)
    {
        scanf("%d%d",&aa,&bb);
        lin[aa].push_back(bb);
        lin[bb].push_back(aa);
    }
    f[0]=size=n;
    getrt(1,rt=0);
    solve(rt);
    cout<<ans<<endl;
}