Codeforces 161.D. Distance in Tree-树分治(点分治，不容斥版)-树上距离为K的点对数量-蜜汁TLE (VK Cup 2012 Round 1)

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

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).

这道题就是求树上距离为K的点对数量。以前写过<=K的点对数量，直接<=K的数量 - <K的数量，讲道理应该也是可以的，但是一直TLE11和TLE17样例。。。

最后换了一种写法，直接求，没有中间-子树的过程，最后过了，有点迷。。。

不容斥版的快，就这样。

代码:

  1 //树分治-点分治
  2 #include<bits/stdc++.h>
  3 using namespace std;
  4 typedef long long ll;
  5 //#pragma GCC optimize(2)
  6 //#define FI(n) FastIO::read(n)
  7 const int inf=1e9+7;
  8 const int maxn=1e5+10;
  9 const int maxm=500+10;
 10 
 11 int head[maxn<<1],tot;
 12 int root,allnode,n,m,k;
 13 bool vis[maxn];
 14 int deep[maxn],dis[maxn],siz[maxn],maxv[maxn];//deep[0]子节点个数(路径长度)，maxv为重心节点
 15 int num[maxm],cnt[maxm];
 16 ll ans=0;
 17 
 18 //namespace FastIO {//读入挂
 19 //    const int SIZE = 1 << 16;
 20 //    char buf[SIZE], obuf[SIZE], str[60];
 21 //    int bi = SIZE, bn = SIZE, opt;
 22 //    int read(char *s) {
 23 //        while (bn) {
 24 //            for (; bi < bn && buf[bi] <= ' '; bi++);
 25 //            if (bi < bn) break;
 26 //            bn = fread(buf, 1, SIZE, stdin);
 27 //            bi = 0;
 28 //        }
 29 //        int sn = 0;
 30 //        while (bn) {
 31 //            for (; bi < bn && buf[bi] > ' '; bi++) s[sn++] = buf[bi];
 32 //            if (bi < bn) break;
 33 //            bn = fread(buf, 1, SIZE, stdin);
 34 //            bi = 0;
 35 //        }
 36 //        s[sn] = 0;
 37 //        return sn;
 38 //    }
 39 //    bool read(int& x) {
 40 //        int n = read(str), bf;
 41 //
 42 //        if (!n) return 0;
 43 //        int i = 0; if (str[i] == '-') bf = -1, i++; else bf = 1;
 44 //        for (x = 0; i < n; i++) x = x * 10 + str[i] - '0';
 45 //        if (bf < 0) x = -x;
 46 //        return 1;
 47 //    }
 48 //};
 49 
 50 inline int read()
 51 {
 52     int x=0;char ch=getchar();
 53     while(ch<'0'||ch>'9')ch=getchar();
 54     while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
 55     return x;
 56 }
 57 
 58 struct node{
 59     int to,next,val;
 60 }edge[maxn<<1];
 61 
 62 void add(int u,int v,int w)//前向星存图
 63 {
 64     edge[tot].to=v;
 65     edge[tot].next=head[u];
 66     edge[tot].val=w;
 67     head[u]=tot++;
 68 }
 69 
 70 void init()//初始化
 71 {
 72     memset(head,-1,sizeof head);
 73     memset(vis,0,sizeof vis);
 74     tot=0;
 75 }
 76 
 77 void get_root(int u,int father)//重心
 78 {
 79     siz[u]=1;maxv[u]=0;
 80     for(int i=head[u];~i;i=edge[i].next){
 81         int v=edge[i].to;
 82         if(v==father||vis[v]) continue;
 83         get_root(v,u);//递归得到子树大小
 84         siz[u]+=siz[v];
 85         maxv[u]=max(maxv[u],siz[v]);//更新u节点的maxv
 86     }
 87     maxv[u]=max(maxv[u],allnode-siz[u]);//保存节点size
 88     if(maxv[u]<maxv[root]) root=u;//更新当前子树的重心
 89 }
 90 
 91 void get_dis(int u,int father)//获取子树所有节点与根的距离
 92 {
 93     if(dis[u]>k) return ;
 94     ans+=num[k-dis[u]];
 95     cnt[dis[u]]++;//计数
 96     for(int i=head[u];~i;i=edge[i].next){
 97         int v=edge[i].to;
 98         if(v==father||vis[v]) continue;
 99         int w=edge[i].val;
100         dis[v]=dis[u]+w;
101         get_dis(v,u);
102     }
103 }
104 
105 void cal(int u,int now)
106 {
107     for(int i=1;i<=k;i++){//初始化，清空
108         num[i]=0;
109     }
110     num[0]=1;
111     for(int i=head[u];~i;i=edge[i].next){
112         int v=edge[i].to;
113         if(vis[v]) continue;
114         for(int j=0;j<=k;j++){//初始化
115             cnt[j]=0;
116         }
117         dis[v]=now;
118         get_dis(v,u);//跑路径
119         for(int j=0;j<=k;j++){
120             num[j]+=cnt[j];//计数
121         }
122     }
123 }
124 
125 void solve(int u)//分治处理
126 {
127     cal(u,1);
128     vis[u]=1;
129     for(int i=head[u];~i;i=edge[i].next){
130         int v=edge[i].to;
131         int w=edge[i].val;
132         if(vis[v]) continue;
133         allnode=siz[v];
134         root=0;
135         get_root(v,u);
136         solve(root);
137     }
138 }
139 
140 int main()
141 {
142 //    FI(n);FI(k);
143     n=read();k=read();
144     init();
145     for(int i=1;i<n;i++){
146         int u,v,w;w=1;
147 //        FI(u);FI(v);
148         u=read();v=read();
149         add(u,v,w);
150         add(v,u,w);
151     }
152     root=0;allnode=n;maxv[0]=inf;
153     get_root(1,0);
154     solve(root);
155     printf("%lld
",ans);
156     return 0;
157 }