E. Close Vertices
You've got a weighted tree, consisting of n vertices. Each edge has a non-negative weight. The length of the path between any two vertices of the tree is the number of edges in the path. The weight of the path is the total weight of all edges it contains.
Two vertices are close if there exists a path of length at most l between them and a path of weight at most w between them. Count the number of pairs of vertices v, u (v < u), such that vertices v and u are close.
Input
The first line contains three integers n, l and w (1 ≤ n ≤ 105, 1 ≤ l ≤ n, 0 ≤ w ≤ 109). The next n - 1 lines contain the descriptions of the tree edges. The i-th line contains two integers pi, wi (1 ≤ pi < (i + 1), 0 ≤ wi ≤ 104), that mean that the i-th edge connects vertex (i + 1)and pi and has weight wi.
Consider the tree vertices indexed from 1 to n in some way.
Output
Print a single integer — the number of close pairs.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64dspecifier.
题目大意
给出一个树，每条边有权值，求经过少于L条边，权值和少于W的路径总数。
题解
点分治裸题，主要是处理答案这里。首先处理以root为根时每个点的到根的边数和权值，然后以边数作为树状数组下标，对于边权值大于要求的-1，查询1-L区间的和就是所求答案。然后又因为将这些边用权值排序，满足单调性，用单调队列进行维护即可。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define int long long
using namespace std;
const int maxn=1e5+10;
const int maxm=1e7+10;
const int INF=0x3f3f3f3f;
int read(){
int x=0,f=1;char ch=getchar();
while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
while(isdigit(ch)){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
int n,ml;
int head[maxn],cnt;
struct edge{int to,nxt,w;}e[maxn*2];
void add(int x,int y,int w){
e[++cnt]=(edge){y,head[x],w};
head[x]=cnt;
}
int vis[maxn];
int size[maxn],son[maxn];
int S,m;
int mx,root;
struct stu{
int w,num;
}dis[maxn];
int c[maxn];
bool cmp(stu a,stu b){return a.w==b.w?a.num<b.num:a.w<b.w;}
void getroot(int x,int f){
size[x]=1;son[x]=0;
for(int i=head[x];i;i=e[i].nxt){
int u=e[i].to;
if(u==f||vis[u])continue;
getroot(u,x);
size[x]+=size[u];
son[x]=max(son[x],size[u]);
}
son[x]=max(son[x],S-size[x]);
if(son[x]<mx){mx=son[x];root=x;}
}
int tot,len,W;
void add(int x,int w){
while(x<=ml){
c[x]+=w;
x+=x&(-x);
}
}
int ask(int x){
int res=0;
while(x>0){
res+=c[x];
x-=x&(-x);
}
return res;
}
void getdis(int x,int f,int v,int t){
if(f==0&&v==0&&t==0){}
else {dis[++tot].w=v,dis[tot].num=t;}
for(int i=head[x];i;i=e[i].nxt){
int u=e[i].to;
if(u==f||vis[u])continue;
getdis(u,x,v+e[i].w,t+1);
}
}
int ans;
int calc(int x,int v,int t){
int res=0;tot=0;
ml=0;
getdis(x,0,v,t);
sort(dis+1,dis+tot+1,cmp);
if(v==0&&t==0)
for(int i=1;i<=tot;i++){
if(dis[i].w<=W&&len>=dis[i].num)res++;
else if(dis[i].w>W)break;
}
int j=tot;
for(int i=1;i<=tot;i++)ml=max(ml,dis[i].num);
for(int i=1;i<=ml;i++)c[i]=0;
for(int i=1;i<=tot;i++)add(dis[i].num,1);
for(int i=1;i<=tot;i++){
while(i<j&&dis[i].w+dis[j].w>W){
add(dis[j].num,-1);
j--;
}
if(i>=j)break;
add(dis[i].num,-1);
res+=ask(min(ml,len-dis[i].num));
}
return res;
}
void dfs(int x){
ans+=calc(x,0,0);
vis[x]=1;
for(int i=head[x];i;i=e[i].nxt){
int u=e[i].to;
if(vis[u])continue;
ans-=calc(u,e[i].w,1);
root=0;S=size[u];mx=INF;getroot(u,0);
dfs(root);
}
}
#undef int
int main()
#define int long long
{
n=read();len=read();W=read();
for(int i=1;i<n;i++){
int y=read(),w=read();
add(i+1,y,w);add(y,i+1,w);
}
S=n;mx=INF;getroot(1,0);
dfs(root);
cout<<ans;
return 0;
}