[BZOJ1468]Tree

Description
给你一棵TREE,以及这棵树上边的距离.问有多少对点它们两者间的距离小于等于K

Input
N(n<=40000) 接下来n-1行边描述管道,按照题目中写的输入 接下来是k

Output
一行,有多少对点之间的距离小于等于k

Sample Input
7
1 6 13
6 3 9
3 5 7
4 1 3
2 4 20
4 7 2
10

Sample Output
5


前置知识:点分治

后置知识:还能有啥后置知识,处理过程中sort+two point扫描即可

/*program from Wolfycz*/
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define inf 0x7f7f7f7f
#define sqr(x) ((x)*(x))
using namespace std;
typedef long long ll;
typedef unsigned int ui;
typedef unsigned long long ull;
inline char gc(){
	static char buf[1000000],*p1=buf,*p2=buf;
	return p1==p2&&(p2=(p1=buf)+fread(buf,1,1000000,stdin),p1==p2)?EOF:*p1++;
}
inline int frd(){
	int x=0,f=1;char ch=gc();
	for (;ch<'0'||ch>'9';ch=gc())	if (ch=='-')    f=-1;
	for (;ch>='0'&&ch<='9';ch=gc())	x=(x<<1)+(x<<3)+ch-'0';
	return x*f;
}
inline int read(){
	int x=0,f=1;char ch=getchar();
	for (;ch<'0'||ch>'9';ch=getchar())	if (ch=='-')	f=-1;
	for (;ch>='0'&&ch<='9';ch=getchar())	x=(x<<1)+(x<<3)+ch-'0';
	return x*f;
}
inline void print(int x){
	if (x<0)    putchar('-'),x=-x;
	if (x>9)	print(x/10);
	putchar(x%10+'0');
}
const int N=4e4;
int pre[(N<<1)+10],now[N+10],child[(N<<1)+10],val[(N<<1)+10];
int size[N+10],dis[N+10],h[N+10];
bool vis[N+10];
int tot,Max,root,K,top;
ll Ans;
void join(int x,int y,int z){pre[++tot]=now[x],now[x]=tot,child[tot]=y,val[tot]=z;}
void insert(int x,int y,int z){join(x,y,z),join(y,x,z);}
void Get_root(int x,int fa,int sz){
	int res=0; size[x]=1;
	for (int p=now[x],son=child[p];p;p=pre[p],son=child[p]){
		if (son==fa||vis[son])	continue;
		Get_root(son,x,sz);
		size[x]+=size[son];
		res=max(res,size[son]);
	}
	res=max(res,sz-size[x]);
	if (res<Max)	Max=res,root=x;
}
void get_dis(int x,int fa){
	h[++top]=dis[x];
	for (int p=now[x],son=child[p];p;p=pre[p],son=child[p]){
		if (son==fa||vis[son])	continue;
		dis[son]=dis[x]+val[p];
		get_dis(son,x);
	}
}
int solve(int x,int v){
	top=0,dis[x]=v;
	get_dis(x,0);
	sort(h+1,h+1+top);
	int l=1,r=top; ll res=0;
	for (;l<=r;l++){
		while (l<=r&&h[l]+h[r]>K)	r--;
		res+=r-l+1;
	}
	return res;
}
void divide(int x){
	vis[x]=1,Ans+=solve(x,0);
	for (int p=now[x],son=child[p];p;p=pre[p],son=child[p]){
		if (vis[son])	continue;
		Ans-=solve(son,val[p]);
		Max=inf,root=0;
		Get_root(son,0,size[son]);
		divide(root);
	}
}
int main(){
	int n=read();
	for (int i=1;i<n;i++){
		int x=read(),y=read(),z=read();
		insert(x,y,z);
	}K=read();
	Max=inf,root=0;
	Get_root(1,0,n);
	divide(root);
	printf("%lld\n",Ans-n);
	return 0;
}
作者:Wolfycz
本文版权归作者和博客园共有,欢迎转载,但必须在文章开头注明原文出处,否则保留追究法律责任的权利
posted @ 2019-01-03 12:08  Wolfycz  阅读(107)  评论(0编辑  收藏  举报