# 洛谷P3066 [USACO12DEC]逃跑的Barn (线段树合并)

It's milking time at Farmer John's farm, but the cows have all run away! Farmer John needs to round them all up, and needs your help in the search.

FJ's farm is a series of N (1 <= N <= 200,000) pastures numbered 1...N connected by N - 1 bidirectional paths. The barn is located at pasture 1, and it is possible to reach any pasture from the barn.

FJ's cows were in their pastures this morning, but who knows where they ran to by now. FJ does know that the cows only run away from the barn, and they are too lazy to run a distance of more than L. For every pasture, FJ wants to know how many different pastures cows starting in that pasture could have ended up in.

Note: 64-bit integers (int64 in Pascal, long long in C/C++ and long in Java) are needed to store the distance values.

* Line 1: 2 integers, N and L (1 <= N <= 200,000, 1 <= L <= 10^18)

* Lines 2..N: The ith line contains two integers p_i and l_i. p_i (1 <= p_i < i) is the first pasture on the shortest path between pasture i and the barn, and l_i (1 <= l_i <= 10^12) is the length of that path.

* Lines 1..N: One number per line, the number on line i is the number pastures that can be reached from pasture i by taking roads that lead strictly farther away from the barn (pasture 1) whose total length does not exceed L.

4 5
1 4
2 3
1 5

3
2
1
1

Cows from pasture 1 can hide at pastures 1, 2, and 4.

Cows from pasture 2 can hide at pastures 2 and 3.

Pasture 3 and 4 are as far from the barn as possible, and the cows can hide there.

#include<bits/stdc++.h>
#define lson tr[now].l
#define rson tr[now].r
#define pii pair<int,long long>
#define mp make_pair
using namespace std;

struct tree
{
int l,r,sum;
}tr[5000020];
vector<pii> g[400010];
int n,cnt,cnt2,rt[400010],deep[400010],q[400010],ans[400010];
long long gg,tmp[400010],dis[400010];
int N=400000;

int push_up(int now)
{
tr[now].sum=tr[lson].sum+tr[rson].sum;
}

int insert(int &now,int l,int r,int pos,int val)
{
if(!now) now=++cnt;
if(l==r)
{
tr[now].sum+=val;
return 0;
}
int mid=(l+r)>>1;
if(pos<=mid)
{
insert(lson,l,mid,pos,val);
}
else
{
insert(rson,mid+1,r,pos,val);
}
push_up(now);
}

int query(int now,int l,int r,int ll,int rr)
{
if(ll<=l&&r<=rr)
{
return tr[now].sum;
}
int mid=(l+r)>>1;
if(rr<=mid)
{
return query(lson,l,mid,ll,rr);
}
else
{
if(mid<ll)
{
return query(rson,mid+1,r,ll,rr);
}
else
{
return query(lson,l,mid,ll,mid)+query(rson,mid+1,r,mid+1,rr);
}
}
}

int merge(int a,int b,int l,int r)
{
if(!a) return b;
if(!b) return a;
if(l==r)
{
tr[a].sum+=tr[b].sum;
return a;
}
int mid=(l+r)>>1;
tr[a].l=merge(tr[a].l,tr[b].l,l,mid);
tr[a].r=merge(tr[a].r,tr[b].r,mid+1,r);
push_up(a);
return a;
}

int dfs(int now,int fa,long long dep)
{
dis[now]=dep;
tmp[++cnt2]=dep;
tmp[++cnt2]=dep+gg;
rt[now]=now;
++cnt;
for(int i=0;i<g[now].size();i++)
{
if(g[now][i].first==fa) continue;
dfs(g[now][i].first,now,dep+g[now][i].second);
}
}

int solve(int now,int fa)
{
insert(rt[now],1,N,deep[now],1);
for(int i=0;i<g[now].size();i++)
{
if(g[now][i].first==fa) continue;
solve(g[now][i].first,now);
merge(rt[now],rt[g[now][i].first],1,N);
}
ans[now]=query(rt[now],1,N,deep[now],q[now]);
}

int init()
{
sort(tmp+1,tmp+cnt2+1);
int tot=unique(tmp+1,tmp+cnt2+1)-tmp-1;
for(int i=1;i<=n;i++)
{
deep[i]=lower_bound(tmp+1,tmp+tot+1,dis[i])-tmp;
}
for(int i=1;i<=n;i++)
{
q[i]=lower_bound(tmp+1,tmp+tot+1,dis[i]+gg)-tmp;
}
}

int main()
{
ios::sync_with_stdio(0);
cin>>n>>gg;
int from;
long long to;
for(int i=2;i<=n;i++)
{
cin>>from>>to;
g[i].push_back(mp(from,to));
g[from].push_back(mp(i,to));
}
dfs(1,0,0);
init();
solve(1,0);
for(int i=1;i<=n;i++)
{
cout<<ans[i]<<endl;
}
}

posted @ 2018-10-17 19:36  Styx-ferryman  阅读(174)  评论(0编辑  收藏  举报