Alyona and a tree CodeForces - 739B (线段树合并)

时间：2021-09-17 11:44:14

大意: 给定有根树, 每个点$x$有权值$a_x$, 对于每个点$x$, 求出$x$子树内所有点$y$, 需要满足$dist(x,y)<=a_y$.

刚开始想错了, 直接打线段树合并了.....因为范围是$long \space long$常数极大, 空间很可能会被卡, 不过竟然过了. 实际上本题每个点对树链上的贡献是单调的, 直接二分就行了

放一下线段树合并代码

#include <iostream>

#include <algorithm>

#include <cstdio>

#include <math.h>

#include <set>

#include <map>

#include <queue>

#include <string>

#include <string.h>

#include <bitset>

#define REP(i,a,n) for(int i=a;i<=n;++i)

#define PER(i,a,n) for(int i=n;i>=a;--i)

#define hr putchar(10)

#define pb push_back

#define lc tr[o].l

#define rc tr[o].r

#define mid ((l+r)>>1)

#define ls lc,l,mid

#define rs rc,mid+1,r

#define x first

#define y second

#define io std::ios::sync_with_stdio(false)

#define endl '\n'

using namespace std;

typedef long long ll;

typedef pair<int,int> pii;

const int P = 1e9+7, INF = 0x3f3f3f3f;

ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}

ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}

ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}

//head

const int N = 2e5+10;

int n, tot, a[N], rt[N], ans[N];

struct _ {int to,w;};

vector<_> g[N];

ll d[N], L, R;

struct {int l,r,sum;} tr[N<<6];

typedef int&& dd;

int merge(int x, int y) {

	if (!x||!y) return x+y;

	tr[x].l=merge(tr[x].l,tr[y].l);

	tr[x].r=merge(tr[x].r,tr[y].r);

	tr[x].sum+=tr[y].sum;

	return x;

}

int query(int o, ll l, ll r, ll ql, ll qr) {

	if (!o||ql<=l&&r<=qr) return tr[o].sum;

	int ans = 0;

	if (mid>=ql) ans+=query(ls,ql,qr);

	if (mid<qr) ans+=query(rs,ql,qr);

	return ans;

}

void update(int &o, ll l, ll r, ll x) {

	if (!o) o=++tot;

	++tr[o].sum;

	if (l==r) return;

	if (mid>=x) update(ls,x);

	else update(rs,x);

}

void dfs(int x) {

	for (auto &&e:g[x]) d[e.to]=d[x]+e.w,dfs(e.to);

	L = min(L, d[x]), R = max(R, d[x]);

	L = min(L, d[x]-a[x]), R = max(R, d[x]-a[x]);

}

void solve(int x) {

	for (auto &&e:g[x]) {

		solve(e.to); rt[x]=merge(rt[x],rt[e.to]);

	}

	ans[x] = query(rt[x],L,R,L,d[x]);

	update(rt[x],L,R,d[x]-a[x]);

}

int main() {

	int &&t = n;

	scanf("%d", &n);

	REP(i,1,n) scanf("%d", a+i);

	REP(i,2,n) {

		int f, w;

		scanf("%d%d", &f, &w);

		g[f].pb({i,w});

	}

	dfs(1),solve(1);

	REP(i,1,n) printf("%d ", ans[i]);hr;

}

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