E. Split the Tree 题解(树上倍增)

题目链接

题目思路

就是用树上倍增首先预处理每个点向上最远跳几步

然后再dfs选取最优的子儿子即可

代码

#include<bits/stdc++.h>
#define fi first
#define se second
#define debug cout<<"I AM HERE"<<endl;
using namespace std;
typedef long long ll;
const int maxn=1e5+5,inf=0x3f3f3f3f,mod=998244353;
const double eps=1e-7;
int n,l;
ll s;
int fa[maxn][25],a[maxn];
ll v[maxn][25];
int f[maxn];
vector<int> g[maxn];
int pr=0;
int dfs(int x){
    int ma=0;
    for(auto nxt:g[x]){
        ma=max(ma,dfs(nxt));
    }
    if(ma==0){
        pr++;
        return f[x]-1;
    }
    return ma-1;
}
signed main(){
    bool flag=1;
    scanf("%d%d%lld",&n,&l,&s);
    for(int i=1;i<=n;i++){
        scanf("%d",&a[i]);
        if(a[i]>s) flag=0;
    }
    if(!flag){
        printf("-1\n");
        return 0;
    }
    for(int i=2;i<=n;i++){
        scanf("%d",&fa[i][0]);
        g[fa[i][0]].push_back(i);
        v[i][0]=a[fa[i][0]];
    }
    for(int j=1;j<=20;j++){
        for(int i=1;i<=n;i++){
            fa[i][j]=fa[fa[i][j-1]][j-1];
            v[i][j]=v[i][j-1]+v[fa[i][j-1]][j-1];
        }
    }
    for(int i=2;i<=n;i++){
        int sz=l-1,now=i;
        ll sum=s-a[i];
        f[i]=1;
        for(int j=20;j>=0;j--){
            if(!fa[now][j]) continue;
            if((1<<j)>sz) continue;
            if(v[now][j]>sum) continue;
            f[i]+=(1<<j);
            sum-=v[now][j];
            sz-=(1<<j);
            now=fa[now][j];
        }
    }
    dfs(1);
    printf("%d\n",pr);
    return 0;
}