[POJ 1741]Tree（树分治）

Description

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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.

Input

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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.

Output

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

Sample Input

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

Sample Output

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8

Source

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LouTiancheng@POJ

Solution

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跨年赛的一道题，树分治入门题（？）
推荐看集训队论文 漆子超：分治算法在树的路径问题中的应用

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#define Max(a,b) (a>b?a:b)
using namespace std;
int n,k,root,cnt,ans,num,nodenum,head[10005],deep[10005],maxv[10005],size[10005];
bool visited[10005];
struct Node
{
    int next,to,w;
}edges[20010];
void add(int u,int v,int w)
{
    edges[cnt].next=head[u];
    edges[cnt].to=v;
    edges[cnt].w=w;
    head[u]=cnt;cnt++;
}
void get_root(int u,int f)
{
    size[u]=1;
    maxv[u]=0;
    for(int i=head[u];~i;i=edges[i].next)
    {
        if(edges[i].to==f||visited[edges[i].to])continue;
        get_root(edges[i].to,u);
        size[u]+=size[edges[i].to];
        maxv[u]=Max(maxv[u],size[edges[i].to]);
    }
    maxv[u]=Max(maxv[u],nodenum-size[u]);
    if(maxv[u]<maxv[root])root=u;
}
void get_deep(int u,int f,int now)
{
    deep[++num]=now;
    for(int i=head[u];~i;i=edges[i].next)
    {
        if(edges[i].to==f||visited[edges[i].to])continue;
        get_deep(edges[i].to,u,now+edges[i].w);
    }
}
int calc(int u,int now)
{
    int res=0;
    num=0;
    get_deep(u,0,now);
    sort(deep+1,deep+1+num);
    for(int l=1,r=num;l<r;)
    {
        if(deep[l]+deep[r]<=k){res+=r-l;l++;}
        else r--;
    }
    return res;
}
void work(int u)
{
    visited[u]=1;
    ans+=calc(u,0);
    for(int i=head[u];~i;i=edges[i].next)
    {
        if(visited[edges[i].to])continue;
        ans-=calc(edges[i].to,edges[i].w);
        root=0;
        nodenum=size[edges[i].to];
        get_root(edges[i].to,u);
        work(root);
    }
}
int main()
{
    while(~scanf("%d%d",&n,&k)&&n!=0)
    {
        cnt=0;ans=0;root=0;maxv[0]=0x3f3f3f3f;
        memset(size,0,sizeof(size));
        memset(visited,0,sizeof(visited));
        memset(head,-1,sizeof(head));
        int u,v,w;
        for(int i=1;i<n;i++)
        {
            scanf("%d%d%d",&u,&v,&w);
            add(u,v,w);
            add(v,u,w);
        }
        get_root(1,0);
        work(root);
        printf("%d\n",ans);
    }
    return 0;
}