POJ 1741:Tree (淀粉质（点分治

Tree

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 22376 Accepted: 7382

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. 

Input

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

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

Sample Input

5 41 2 31 3 11 4 23 5 10 0

Sample Output

8

Source

LouTiancheng@POJ

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题意：给定一棵无根树，有边权，求有多少个点对距离小于k。

题解：很久以前做了这道题，当时学习了点分治，今天再来看看，具体可以看一篇论文《分治算法在树的路径中的应用》，实际上每次分治就是考虑是否经过根节点，然后一层层分治下去，对于这道题，要不断的寻找重心作为根然后分治，就解决了。

贴上代码：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#define inf 0x7fffffffusing namespace std;int read(){int x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}return x*f;}int n,K,cnt,sum,ans,root;int head[10005],deep[10005],d[10005],f[10005],son[10005];bool vis[10005];struct data{int to,next,v;}e[20005];inline void ins(int u,int v,int w){e[++cnt].to=v;e[cnt].next=head[u];head[u]=cnt;e[cnt].v=w;}inline void insert(int u,int v,int w){ins(u,v,w);ins(v,u,w);}void getroot(int x,int fa){son[x]=1;f[x]=0;for(int i=head[x];i;i=e[i].next){if(e[i].to==fa||vis[e[i].to])continue;getroot(e[i].to,x);son[x]+=son[e[i].to];f[x]=max(f[x],son[e[i].to]);}f[x]=max(f[x],sum-son[x]);if(f[x]<f[root])root=x;}void getdeep(int x,int fa){deep[++deep[0]]=d[x];for(int i=head[x];i;i=e[i].next){if(e[i].to==fa||vis[e[i].to])continue;d[e[i].to]=d[x]+e[i].v;getdeep(e[i].to,x);}}int cal(int x,int now){d[x]=now;deep[0]=0;getdeep(x,0);sort(deep+1,deep+deep[0]+1);int t=0,l,r;for(l=1,r=deep[0];l<r;){if(deep[l]+deep[r]<=K){t+=r-l;l++;}        else r--;    }return t;}void work(int x){ans+=cal(x,0);vis[x]=1;for(int i=head[x];i;i=e[i].next){if(vis[e[i].to])continue;ans-=cal(e[i].to,e[i].v);sum=son[e[i].to];root=0;getroot(e[i].to,root);work(root);}}int main(){while(true){ans=0;root=0;cnt=0;memset(vis,0,sizeof(vis));memset(head,0,sizeof(head));    n=read();K=read();if(n==0)break;    for(int i=1;i<n;i++)     {    int u=read(),v=read(),w=read();    insert(u,v,w);    }    sum=n;f[0]=inf;     getroot(1,0);    work(root);    printf("%d\n",ans);    }return 0;}