题目简述
一个\(N\)个节点的树，有\(M\)个炸弹分布在一些节点上，有各自的威力，随着其他点距离增大对其他点的伤害呈等差减小，直至为0
问每个点受到的伤害
题解
QAQ考场代码没处理好有些炸弹威力很大这个事实，，数组爆掉。。。
AC算法直接变暴力分，，，
点分治即可
我是每次将子树内所有的炸弹统计到根来，再用一个差分数组求出各个深度受到的伤害，累加入每个节点的答案
但是由于可能会出现伤害来自同一个子树的情况，我们再对每个子树做一遍撤销
常数略大，，经玄学优化强行卡入时限
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<vector>
#include<cmath>
#include<ctime>
#define LL long long int
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
using namespace std;
const int maxn = 500005,maxm = 600005,INF = 100000000;
inline int read(){
int out = 0; char c = getchar();
while (c < 48 || c > 57) c = getchar();
while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
return out;
}
inline void write(LL x){
LL tmp = 0; int cnt = 0;
while (x) tmp = (tmp << 3) + (tmp << 1) + x % 10,x /= 10,cnt++;
while (cnt--) putchar(tmp % 10 + 48),tmp /= 10;
putchar(‘\n‘);
}
int h[maxn],ne = 2,n,m;
struct EDGE{int to,nxt;}ed[maxn * 2];
int F[maxn],Siz[maxn],rt,sum,vis[maxn];
int siz[maxn],st[maxm],top,d[maxn],fa[maxn],nd[maxn],ndi;
int md;
LL ans[maxn],D[maxn];
vector<int> power[maxn];
inline void build(int u,int v){
ed[ne] = (EDGE){v,h[u]}; h[u] = ne++;
ed[ne] = (EDGE){u,h[v]}; h[v] = ne++;
}
void getRT(int u){
Siz[u] = 1 + power[u].size();
F[u] = 0;
Redge(u) if (!vis[to = ed[k].to] && to != fa[u]){
fa[to] = u;
getRT(to);
Siz[u] += Siz[to];
F[u] = max(F[u],Siz[to]);
}
F[u] = max(F[u],sum - Siz[u]);
if (F[u] < F[rt]) rt = u;
}
void dfs1(int u){
Siz[u] = 1 + power[u].size(); siz[u] = 1;
nd[++ndi] = u;
md = max(md,d[u]);
Redge(u) if (!vis[to = ed[k].to] && to != fa[u]){
fa[to] = u; d[to] = d[u] + 1;
dfs1(to);
Siz[u] += Siz[to];
siz[u] += siz[to];
}
}
void dfs4(int u){
nd[++ndi] = u;
for (int j = 0; j < power[u].size(); j++){
if (power[u][j] > d[u] + 1) st[++top] = power[u][j] - d[u] - 1;
}
Redge(u) if (!vis[to = ed[k].to] && to != fa[u]){
dfs4(to);
}
}
void dfs5(int u){
ans[u] -= D[d[u]];
Redge(u) if (!vis[to = ed[k].to] && to != fa[u]){
dfs5(to);
}
}
void cal(int u){
top = 0;
for (int j = 0; j < power[u].size(); j++)
if (power[u][j] > 2) st[++top] = power[u][j] - 2;
for (int i = 0; i <= siz[u]; i++) D[i] = 0;
ndi = 0;
Redge(u) if (!vis[to = ed[k].to] && to != fa[u]){
dfs4(to);
}
for (int i = 1; i <= top; i++){
if (st[i] <= md){
D[1] += st[i];
D[2] += -1 - st[i];
D[st[i] + 2] += 1;
}
else {
D[1] += st[i];
D[2] += -1 - st[i];
}
}
for (int i = 1; i <= siz[u]; i++) D[i] += D[i - 1];
for (int i = 1; i <= siz[u]; i++) D[i] += D[i - 1];
ans[u] -= D[1];
for (int i = 1; i <= ndi; i++) ans[nd[i]] -= D[d[nd[i]]];
}
void solve(int u){
vis[u] = true; siz[u] = 1; Siz[u] = 1 + power[u].size();
ndi = 0; md = 0;
Redge(u) if (!vis[to = ed[k].to]){
fa[to] = u; d[to] = 1;
dfs1(to);
siz[u] += siz[to];
Siz[u] += Siz[to];
}
md += 2;
if (Siz[u] == siz[u]) return;
top = 0;
for (int j = 0; j < power[u].size(); j++) st[++top] = power[u][j];
for (int i = 1; i <= ndi; i++){
int v = nd[i];
for (int j = 0; j < power[v].size(); j++){
if (power[v][j] > d[v]) st[++top] = power[v][j] - d[v];
}
}
for (int i = 0; i <= siz[u]; i++) D[i] = 0;
for (int i = 1; i <= top; i++){
if (st[i] <= md){
D[0] += st[i];
D[1] += -1 - st[i];
D[st[i] + 1] += 1;
}
else {
D[0] += st[i];
D[1] += -1 - st[i];
}
}
for (int i = 1; i <= siz[u]; i++) D[i] += D[i - 1];
for (int i = 1; i <= siz[u]; i++) D[i] += D[i - 1];
ans[u] += D[0];
for (int i = 1; i <= ndi; i++) ans[nd[i]] += D[d[nd[i]]];
Redge(u) if (!vis[to = ed[k].to]){
if (siz[to] == Siz[to]) continue;
cal(to);
}
Redge(u) if (!vis[to = ed[k].to]){
if (siz[to] == Siz[to]) continue;
sum = Siz[to]; F[rt = 0] = INF;
getRT(to);
solve(rt);
}
}
int main(){
//double t = clock();
//freopen("1.in","r",stdin);
//freopen("a.out","w",stdout);
n = read(); m = read();
int pos,w;
for (int i = 2; i <= n; i++) build(i,read());
for (int i = 1; i <= m; i++){
pos = read(); w = read();
power[pos].push_back(w);
}
F[rt = 0] = INF; sum = n + m;
getRT(1);
solve(rt);
for (int i = 1; i <= n; i++) write(ans[i]);
//cerr << (clock() - t) / CLOCKS_PER_SEC << ‘s‘ << endl;
return 0;
}
