Description

给你一棵TREE,以及这棵树上边的距离.问有多少对点它们两者间的距离小于等于K

Input

N（n<=40000） 接下来n-1行边描述管道，按照题目中写的输入 接下来是k

Output

一行，有多少对点之间的距离小于等于k

Sample Input

7

1 6 13

6 3 9

3 5 7

4 1 3

2 4 20

4 7 2

10

Sample Output

5

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前置知识：

后置知识：还能有啥后置知识，处理过程中sort+two point扫描即可

/*program from Wolfycz*/#include
    
     #include
     
      #include
      
       #include
       
        #include
        
         #define inf 0x7f7f7f7f#define sqr(x) ((x)*(x))using namespace std;typedef long long ll;typedef unsigned int ui;typedef unsigned long long ull;inline char gc(){    static char buf[1000000],*p1=buf,*p2=buf;    return p1==p2&&(p2=(p1=buf)+fread(buf,1,1000000,stdin),p1==p2)?EOF:*p1++;}inline int frd(){    int x=0,f=1;char ch=gc();    for (;ch<'0'||ch>'9';ch=gc())   if (ch=='-')    f=-1;    for (;ch>='0'&&ch<='9';ch=gc()) x=(x<<1)+(x<<3)+ch-'0';    return x*f;}inline int read(){    int x=0,f=1;char ch=getchar();    for (;ch<'0'||ch>'9';ch=getchar())  if (ch=='-')    f=-1;    for (;ch>='0'&&ch<='9';ch=getchar())    x=(x<<1)+(x<<3)+ch-'0';    return x*f;}inline void print(int x){    if (x<0)    putchar('-'),x=-x;    if (x>9)    print(x/10);    putchar(x%10+'0');}const int N=4e4;int pre[(N<<1)+10],now[N+10],child[(N<<1)+10],val[(N<<1)+10];int size[N+10],dis[N+10],h[N+10];bool vis[N+10];int tot,Max,root,K,top;ll Ans;void join(int x,int y,int z){pre[++tot]=now[x],now[x]=tot,child[tot]=y,val[tot]=z;}void insert(int x,int y,int z){join(x,y,z),join(y,x,z);}void Get_root(int x,int fa,int sz){    int res=0; size[x]=1;    for (int p=now[x],son=child[p];p;p=pre[p],son=child[p]){        if (son==fa||vis[son])  continue;        Get_root(son,x,sz);        size[x]+=size[son];        res=max(res,size[son]);    }    res=max(res,sz-size[x]);    if (res
         
          K) r--; res+=r-l+1; } return res;}void divide(int x){ vis[x]=1,Ans+=solve(x,0); for (int p=now[x],son=child[p];p;p=pre[p],son=child[p]){ if (vis[son]) continue; Ans-=solve(son,val[p]); Max=inf,root=0; Get_root(son,0,size[son]); divide(root); }}int main(){ int n=read(); for (int i=1;i