[POJ1741]Tree

题目描述 Description

 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. 

输入描述 Input Description

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. 

输出描述 Output Description

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

样例输入 Sample Input

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

样例输出 Sample Output

8

数据范围及提示 Data Size & Hint

 

一道比较裸的点分治题。对于每一棵树，先找到他的重心，然后算出他的所有子孙到他的dis,统一放在dis数组中，下面只需在数组中找数对满足和为k就好了，这个能在O（n）的时间内解决。但是这样会有一个问题。对于在同一棵子树上的点，路径并不是从这个点跑到重心，再跑下来，于是我们需要去重。去重之后对重心打上标记，表示以后不能再用这个点了，之后递归的处理每一个子树。一道破题调了好久，后来发现是在getd函数中传参数时出了BUG，看来以后函数的参数还是要有顺序的！

#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<queue>
using namespace std;
typedef long long LL;
#define Pi acos(-1.0)
#define mem(a,b) memset(a,b,sizeof(a))
inline int read()
{
    int x=0,f=1;char c=getchar();
    while(!isdigit(c)){if(c=='-')f=-1;c=getchar();}
    while(isdigit(c)){x=x*10+c-'0';c=getchar();}
    return x*f;
}
const int maxn=10010;
struct Edge
{
    int u,v,w,next;
    Edge() {}
    Edge(int _1,int _2,int _3,int _4) : u(_1),v(_2),w(_3),next(_4) {}
}e[2*maxn];
int first[maxn],n,k,a,b,c,ans,size[maxn],masize[maxn],now_size,root,dis[maxn],end;
bool vis[maxn]; 
void addEdge(int i,int a,int b,int c)
{
    e[i]=Edge(a,b,c,first[a]);
    first[a]=i;
}
void gets(int u,int pa)
{
    size[u]=1;
    masize[u]=0;
    for(int i=first[u];i!=-1;i=e[i].next)
        if(!vis[e[i].v] && e[i].v!=pa)
        {
            gets(e[i].v,u);
            size[u]+=size[e[i].v];
            masize[u]=max(size[e[i].v],masize[u]);
        }
}
void getr(int r,int u,int pa)
{
    masize[u]=max(masize[u],size[r]-size[u]);
    if(masize[u]<now_size)now_size=masize[u],root=u;
    for(int i=first[u];i!=-1;i=e[i].next)
        if(!vis[e[i].v] && e[i].v!=pa)getr(r,e[i].v,u);
}
void getd(int u,int pa,int d)
{
    dis[end++]=d;
    for(int i=first[u];i!=-1;i=e[i].next)
        if(!vis[e[i].v] && e[i].v!=pa)getd(e[i].v,u,d+e[i].w);
}
int calc(int u,int d)
{
    end=0;
    getd(u,-1,d);
    int ret=0,l=0,r=end-1;
    sort(dis,dis+end); 
    while(l<r)
    {
        while(dis[l]+dis[r]>k && l<r)r--;
        ret+=(r-l);l++;
    }
    return ret;
}
void dfs(int u)
{
    now_size=n;
    gets(u,-1);
    getr(u,u,-1);
    ans+=calc(root,0);
    vis[root]=1;
    for(int i=first[root];i!=-1;i=e[i].next)
        if(!vis[e[i].v])
        {
            ans-=calc(e[i].v,e[i].w);
            dfs(e[i].v);
        }
    return;
}
void init(){mem(first,-1);mem(size,0);mem(masize,0);mem(vis,0);mem(dis,0);ans=0;}
int main()
{
    while(scanf("%d%d",&n,&k)!=EOF)
    {
        if(n==0 && k==0)break;
        init();
        for(int i=0;i<n-1;i++)
        {
            a=read();b=read();c=read();
            addEdge(i*2,a,b,c);addEdge(i*2+1,b,a,c);
        }
        dfs(1);
        printf("%d
",ans);    
    }
    return 0;
}