Codeforces 1059E. Split the Tree

题目：http://codeforces.com/problemset/problem/1059/E

用倍增可以在nlog内求出每个节点占用一个sequence 时最远可以向父节点延伸到的节点，对每个节点作为sequence 的最后一个元素向上延伸时，将节点的父节点属性合并（类似于并查集的操作），

存在优先队列里保证每次先操作dep最大的节点

#include<iostream>
#include<cstdio> 
#include<cmath>
#include<queue>
#include<vector>
#include<string.h>
#include<cstring>
#include<algorithm>
#include<set>
#include<map>
#include<fstream>
#include<cstdlib>
#include<ctime>
#include<list>
#include<climits>
#include<bitset>
#include<random>
#include <ctime>
#include <cassert>
#include <complex>
#include <cstring>
#include <chrono>
using namespace std;
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fopen freopen("input.in", "r", stdin);freopen("output.in", "w", stdout);
#define left asfdasdasdfasdfsdfasfsdfasfdas1
#define set asfdasdasdfasdfsdfasfsdfasfdas2
#define tan asfdasdasdfasdfasfdfasfsdfasfdas
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
typedef long long ll;
typedef unsigned int un;
const int desll[4][2]={{0,1},{0,-1},{1,0},{-1,0}};
const int mod=1e9+7;
const int maxn=1e5+1;
const int maxm=1e5;
const double eps=1e-8;
const int csize=2;
int n,k,m,ar[maxn];
int f[maxn],tan[maxn][18],v[maxn],in[maxn],mark;
ll ss[maxn][18];
bool ma[maxn],que[maxn];
priority_queue<pair<int,int> > qu;
vector<int> ve[maxn];
int dep[maxn];
void dfs(int u,int pre)
{
    if(pre>=0)dep[u]=dep[pre]+1;
    for(int i=0;i<ve[u].size();i++){
        int v=ve[u][i];
        if(v==pre)continue;
        dfs(v,u);
    }
}
int main()
{
    int l;
    ll s,mx=0;f[1]=0;
    scanf("%d%d%I64d",&n,&l,&s);
    memset(in,0,sizeof(in));
    memset(tan,0,sizeof(tan));
    memset(ma,0,sizeof(ma));
    memset(que,0,sizeof(que));
    for(int i=1;i<=n;i++)scanf("%d",&ar[i]),mx=max(mx,1LL*ar[i]);
    for(int i=2;i<=n;i++)scanf("%d",&f[i]),in[f[i]]++,ve[f[i]].push_back(i);
    dep[1]=1;
    dfs(1,-1);
    if(mx>s){
        printf("-1\n");
        return 0;
    }
    for(int i=1;i<=n;i++){
        tan[i][0]=f[i];
        ss[i][0]=ar[f[i]];
    }
    for(int j=1;j<18;j++){//倍增预处理
        for(int i=1;i<=n;i++){
            if(i + (1<<j) >n)break;
            tan[i][j]=tan[tan[i][j-1]][j-1];
            ss[i][j] = ss[i][j-1]+ss[tan[i][j-1]][j-1];
        }
    }
    for(int i=1;i<=n;i++){
        ll lmid=l-1,smid=s-ar[i];
        int ins=17,x=i;
        while(ins>=0){
            if(tan[x][ins]==0)ins--;
            else if((1<<ins) > lmid)ins--;
            else if(ss[x][ins]<=smid){
                smid-=ss[x][ins];
                lmid -= (1<<ins);
                x=tan[x][ins];
            }
            else ins--;
        }
        v[i]=x;//v[i]代表i节点作为sequence尾节点时最远可以向上延伸到的节点
    }
    for(int i=1;i<=n;i++){
        if(v[i]==0)exit(0);
    }
    for(int i=1;i<=n;i++){
        if(in[i]==0){
            qu.push(make_pair(dep[i],i));
            que[i]=1;
        }
    }
    int ans=0;
    while(qu.size()){
        int u=qu.top().second;qu.pop();
        if(ma[u])continue;//如果已经标记过，代表有其他节点可以延伸到此节点，跳过即可
        mark=v[u];
        int mid=f[u],pre=u;
        while(dep[mid]>dep[mark]){//对ma数组进行标记，合并f数组，类似于并查集的操作
            ma[pre]=1;
            f[pre]=mark;
            pre=mid;
            mid=f[mid];
        }
        ma[pre]=1;
        ma[mark]=1;
        if(pre!=mark)f[pre]=mark;
        ans++;
        //cout<<mark<<" "<<f[mark]<<" "<<ma[f[mark]]<<" "<<que[f[mark]]<<endl;
        if(f[mark]>0 && !ma[f[mark]] && !que[f[mark]]){
            qu.push(make_pair(dep[f[mark]],f[mark]));
            que[f[mark]]=1;
        }
    }
    printf("%d\n",ans);

    return 0;
}