子树操作, dfs序即可.然后计算<=L就直接在可持久化线段树上查询
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#include<bits/stdc++.h>
using namespace std;
#define M(l, r) (((l) + (r)) >> 1)
const int maxn = 200009;
typedef long long ll;
inline ll readll() {
char c = getchar();
for(; !isdigit(c); c = getchar());
ll ans = 0;
for(; isdigit(c); c = getchar())
ans = ans * 10 + c - '0';
return ans;
}
inline int readint() {
char c = getchar();
for(; !isdigit(c); c = getchar());
int ans = 0;
for(; isdigit(c); c = getchar())
ans = ans * 10 + c - '0';
return ans;
}
struct HASH {
ll id[maxn];
int N;
HASH() {
N = 0;
}
void add(ll v) {
id[N++] = v;
}
void work() {
sort(id, id + N);
N = unique(id, id + N) - id;
id[N] = 0x7fffffffffffffff;
}
int hash(ll v) {
return lower_bound(id, id + N, v) - id;
}
int _hash(ll v) {
return upper_bound(id, id + N, v) - id;
}
} h;
struct Node {
Node *l, *r;
int s;
} pool[maxn * 20], *pt = pool, *null, *root[maxn];
void set_null() {
null = pt++;
null->l = null->r = null;
null->s = 0;
}
int p;
Node* modify(Node* t, int l, int r) {
Node* h = pt++;
h->s = t->s + 1;
if(r > l) {
int m = M(l, r);
if(p <= m) {
h->l = modify(t->l, l, m);
h->r = t->r;
} else {
h->l = t->l;
h->r = modify(t->r, m + 1, r);
}
}
return h;
}
int query(int __l, int __r, int v) {
Node *_l = __l ? root[__l - 1] : null, *_r = root[__r];
int L = 1, R = h.N, ans = 0;
while(L < R) {
int m = M(L, R);
if(R <= v) {
ans +=_r->s - _l->s;
break;
}
if(m <= v) {
ans += _r->l->s - _l->l->s;
_l = _l->r; _r = _r->r; L = m + 1;
} else {
_l = _l->l; _r = _r->l; R = m;
}
}
return ans;
}
struct edge {
int to;
ll dist;
edge(int _t, ll _d):to(_t), dist(_d) {}
};
ll seq[maxn], L;
int _L[maxn], _R[maxn], cur = -1, N;
vector<edge> G[maxn];
void dfs(int x, ll d, int fa) {
seq[_L[x] = ++cur] = d;
for(vector<edge>::iterator e = G[x].begin(); e != G[x].end(); e++)
if(e->to != fa) dfs(e->to, d + e->dist, x);
_R[x] = cur;
}
void init() {
N = readint(); L = readll();
for(int i = 1; i < N; i++) {
int p = readint() - 1; ll d = readll();
G[p].push_back(edge(i, d));
G[i].push_back(edge(p, d));
}
set_null();
}
void work() {
dfs(0, 0, -1);
for(int i = 0; i < N; i++)
h.add(seq[i]);
h.work();
Node* fa = null;
for(int i = 0; i < N; i++) {
p = h.hash(seq[i]) + 1;
fa = root[i] = modify(fa, 1, h.N);
}
for(int i = 0; i < N; i++)
printf("%d\n", query(_L[i], _R[i], h._hash(L + seq[_L[i]])));
}
int main() {
init();
work();
return 0;
}
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3011: [Usaco2012 Dec]Running Away From the Barn
Time Limit: 10 Sec Memory Limit: 128 MB
Submit: 136 Solved: 74
[Submit][Status][Discuss]Description
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.
给出以1号点为根的一棵有根树，问每个点的子树中与它距离小于l的点有多少个。
Input
* 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.
Output
* 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.
Sample Input
4 5
1 4
2 3
1 5
Sample Output
3
2
1
1
OUTPUT DETAILS: 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.
HINT
Source
