学步园

Give a tree with n vertices,each edge has a length(positive integer less than 1001).

Define dist(u,v)=The min distance between node u and v.
Give an integer k,for every pair (u,v) of vertices is called valid if and only if dist(u,v) not exceed k.

Write a program that will count how many pairs which are valid for a given tree.

The input contains several test cases. The first line of each test case contains two integers n, k. (n<=10000) The following n-1 lines each contains three integers u,v,l, which means there is an edge between node u and v of length
l.
The last test case is followed by two zeros.

For each test case output the answer on a single line.

5 4
1 2 3
1 3 1
1 4 2
3 5 1
0 0

8

题意：求满足dis（i，j）<=k，即距离小于等于k的点对的数目

题解：对树的点进行分治，用O（n）的算法处理出对于某个点到其他节点的距离，然后O（nlogn）排序，显然再用O（n）的算法就可以算出某2个距离和小于等于k的结点对数（包含在同一棵子树的2个结点对），然后就分别用O（n）遍历子树，减去所有在同一棵子树的结点对数，这样就完成了一次分治。对于树所要删除进行分治的点，最好选择该树（子树）的重心。。。。本人表达捉急。。。详细见代码。。。附带，总时间复杂度为O（nlogn）

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#define MAXN 10008
struct edge{
    int to,len,next;
}p[MAXN*2];
int head[MAXN],son[MAXN],mark[MAXN];
int d[MAXN],dis[MAXN],cou;
int all,root,rsize,n,m,ans;
int MAX(int x,int y){ return x>y?x:y; }
int MIN(int x,int y){ return x<y?x:y; }
int cmp(const void *a,const void *b)
{
    return *(int *)a<*(int *)b?-1:1;
}
void add(int x,int y,int len)
{
    p[all].to=y;
    p[all].len=len;
    p[all].next=head[x];
    head[x]=all++;
}
void getroot(int x,int f)
{
    int res=0,i;

    son[x]=0;
    for(i=head[x];i!=-1;i=p[i].next)
    {
        if(mark[p[i].to]||p[i].to==f) continue;
        getroot(p[i].to,x);
        son[x]+=son[p[i].to];
        res=MAX(res,son[p[i].to]);
    }
    son[x]++;
    res=MAX(res,n-son[x]);
    if(rsize>res) rsize=res,root=x;
}
void getdis(int x,int f)
{
    int i;

    d[cou++]=dis[x];
    for(i=head[x];i!=-1;i=p[i].next)
    {
        if(mark[p[i].to]||p[i].to==f) continue;
        dis[p[i].to]=dis[x]+p[i].len;
        getdis(p[i].to,x);
    }
}
int mycount(int x,int len)
{
    int res=0,l,r;

    dis[x]=len;
    cou=0;
    getdis(x,-1);
    qsort(d,cou,sizeof(d[0]),cmp);
    for(l=0,r=cou-1;l<r;)
    {
        if(d[l]+d[r]<=m) res+=r-l++;
        else r--;
    }

    return res;
}
void solve(int x)
{
    int i;

    ans+=mycount(x,0);
    mark[x]=1;
    for(i=head[x];i!=-1;i=p[i].next)
    {
        if(mark[p[i].to]) continue;
        ans-=mycount(p[i].to,p[i].len);
        root=rsize=n=son[x];
        getroot(p[i].to,-1);
        solve(root);
    }
}
int main()
{
    int x,y,z,i;

    while(scanf("%d%d",&n,&m),n+m)
    {
        memset(head,-1,sizeof(head));
        memset(mark,0,sizeof(mark));
        for(all=i=0;i<n-1;i++)
        {
            scanf("%d%d%d",&x,&y,&z);
            add(x,y,z);  add(y,x,z);
        }
        root=rsize=n,ans=0;
        getroot(1,-1);
        solve(root);
        printf("%d\n",ans);
    }

    return 0;
}