poj1741 Tree

Tree

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 15927		Accepted: 5194

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

Sample Output

8

Source

LouTiancheng@POJ

树的点分治裸题

树的点分治的思想就是对于一棵树，先找出树的重心，然后统计不同子树间的答案(同时也要减去同一棵子树中多统计的答案)，再递归每一棵子树。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#define F(i,j,n) for(int i=j;i<=n;i++)
#define D(i,j,n) for(int i=j;i>=n;i--)
#define ll long long
#define maxn 10005
using namespace std;
int n,k,cnt,tot,mn,root,ans;
int head[maxn],sz[maxn],mx[maxn],dis[maxn];
bool vst[maxn];
struct edge_type{int next,to,w;}e[maxn*2];
inline 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;
}
inline void add_edge(int x,int y,int z)
{
	e[++cnt]=(edge_type){head[x],y,z};head[x]=cnt;
	e[++cnt]=(edge_type){head[y],x,z};head[y]=cnt;
}
inline void dfssize(int x,int fa)
{
	sz[x]=1;
	mx[x]=0;
	for(int i=head[x];i;i=e[i].next)
	{
		int y=e[i].to;
		if (y!=fa&&!vst[y])
		{
			dfssize(y,x);
			sz[x]+=sz[y];
			mx[x]=max(mx[x],sz[y]);
		}
	}
}
inline void dfsroot(int rt,int x,int fa)
{
	mx[x]=max(mx[x],sz[rt]-sz[x]);
	if (mx[x]<mn) mn=mx[x],root=x;
	for(int i=head[x];i;i=e[i].next)
	{
		int y=e[i].to;
		if (y!=fa&&!vst[y]) dfsroot(rt,y,x);
	}
}
inline void dfsdis(int x,int d,int fa)
{
	dis[++tot]=d;
	for(int i=head[x];i;i=e[i].next)
	{
		int y=e[i].to;
		if (y!=fa&&!vst[y]) dfsdis(y,d+e[i].w,x);
	}
}
inline int calc(int x,int d)
{
	int ret=0;
	tot=0;
	dfsdis(x,d,0);
	sort(dis+1,dis+tot+1);
	int i=1,j=tot;
	while (i<j)
	{
		while (dis[i]+dis[j]>k&&i<j) j--;
		ret+=j-i;
		i++;
	}
	return ret;
}
inline void dfs(int x)
{
	mn=n;
	dfssize(x,0);
	dfsroot(x,x,0);
	ans+=calc(root,0);
	vst[root]=true;
	for(int i=head[root];i;i=e[i].next)
	{
		int y=e[i].to;
		if (!vst[y])
		{
			ans-=calc(y,e[i].w);
			dfs(y);
		}
	}
}
int main()
{
	n=read();k=read();
	while (n)
	{
		memset(head,0,sizeof(head));
		memset(vst,false,sizeof(vst));
		cnt=ans=0;
		F(i,1,n-1)
		{
			int x=read(),y=read(),z=read();
			add_edge(x,y,z);
		}
		dfs(1);
		printf("%d\n",ans);
		n=read();k=read();
	}
	return 0;
}