雅礼培训4.3 Problem A 【点分治】

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;
}

原文地址：https://www.cnblogs.com/Mychael/p/8711249.html