poj 1741 tree

最新推荐文章于 2020-01-18 17:11:33 发布

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最新推荐文章于 2020-01-18 17:11:33 发布

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分类专栏：点分治

本文链接：https://blog.youkuaiyun.com/chai_jing/article/details/72636326

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本文详细解析了一道经典的点分治算法题目，介绍了如何通过点分治算法来解决树上两点间距离不超过特定阈值的有效点对计数问题。文章提供了完整的代码实现，并通过样例输入输出展示了算法的应用。

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Tree

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 22255		Accepted: 7328

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

Sample Output

Source

LouTiancheng@POJ

【分析】

点分治

题解传送门

【代码】

//poj 1741 tree 
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#define ll long long
#define M(a) memset(a,0,sizeof a)
#define fo(i,j,k) for(i=j;i<=k;i++)
using namespace std;
const int mxn=10005;
bool vis[mxn];
int n,m,T,mn,cnt,ans,num,root;
int head[mxn],dis[mxn],size[mxn],mx[mxn];  //mx:以u为根的子树中的最大size 
struct edge {int to,w,next;} f[mxn<<1];
inline void add(int u,int v,int w)
{
	f[++cnt].to=v,f[cnt].w=w,f[cnt].next=head[u],head[u]=cnt;
}
inline void dfssize(int u,int fa)
{
	size[u]=1,mx[u]=0;
	for(int i=head[u];i;i=f[i].next)
	{
		int v=f[i].to;
		if(vis[v] || v==fa) continue;
		dfssize(v,u);
		size[u]+=size[v];
		mx[u]=max(mx[u],size[v]);
	}
}
inline void dfsroot(int r,int u,int fa)
{
	mx[u]=max(mx[u],size[r]-size[u]);
	if(mx[u]<mn) mn=mx[u],root=u;
	for(int i=head[u];i;i=f[i].next)
	{
		int v=f[i].to;
		if(vis[v] || v==fa) continue;
		dfsroot(r,v,u);
	}
}
inline void dfsdis(int w,int u,int fa)
{
	dis[++num]=w;
	for(int i=head[u];i;i=f[i].next)
	{
		int v=f[i].to;
		if(vis[v] || v==fa) continue;
		dfsdis(w+f[i].w,v,u);
	}
}
inline int calc(int u,int d)
{
	int res=0;
	num=0;
	dfsdis(d,u,0);
	sort(dis+1,dis+num+1);
	int i=1,j=num;
	while(i<j)
	{
		while(dis[i]+dis[j]>m && j>i) j--;
		res+=j-i;
		i++;
	}
	return res;
}
inline void dfs(int u)
{
	mn=n;
	dfssize(u,0);
	dfsroot(u,u,0);
	ans+=calc(root,0);
	vis[root]=1;
	for(int i=head[root];i;i=f[i].next)
	{
		int v=f[i].to;
		if(vis[v]) continue;
		ans-=calc(v,f[i].w);
		dfs(v);
	}
}
int main()
{
	int i,j,u,v,w;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		if(!n && !m) return 0;
		M(head),M(vis),cnt=ans=0;
		fo(i,2,n)
		{
			scanf("%d%d%d",&u,&v,&w);
			add(u,v,w),add(v,u,w);
		}
		dfs(1);
		printf("%d\n",ans);
	}
	return 0;
}
/*
5 4
1 2 3
1 3 1
1 4 2
3 5 1
0 0
*/