【Nowcoder】城市网络 倍增 + DP

传送门

分析

小丑竟是我自己系列
一开始在觉得主席树能写，从叶子结点到根节点维护一下递增序列就可以了，后来发现好像写不了
考虑$dp$关系，设$f[u][x]$表示$u$结点向上购买$x$次所能到达的节点，所以状态转移方程就是$f[u][x]=f[f[u][y][x-y]$，我们用倍增的思想去转化这个关系即可

代码

#pragma GCC optimize(3)
#include <bits/stdc++.h>
#define debug(x) cout<<#x<<":"<<x<<endl;
#define dl(x) printf("%lld\n",x);
#define di(x) printf("%d\n",x);
#define _CRT_SECURE_NO_WARNINGS
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
typedef vector<int> VI;
const int INF = 0x3f3f3f3f;
const int N = 2e5 + 10,M = N * 2;
const ll mod = 1000000007;
const double eps = 1e-9;
const double PI = acos(-1);
template<typename T>inline void read(T &a) {
    char c = getchar(); T x = 0, f = 1; while (!isdigit(c)) {if (c == '-')f = -1; c = getchar();}
    while (isdigit(c)) {x = (x << 1) + (x << 3) + c - '0'; c = getchar();} a = f * x;
}
int gcd(int a, int b) {return (b > 0) ? gcd(b, a % b) : a;}
int fa[N][25],dep[N],a[N],q[N];
int n,m;
int h[N],e[M],ne[M],idx;

void add(int x,int y){
    ne[idx] = h[x],e[idx] = y,h[x] = idx++;
}

void dfs(int u,int f){
    dep[u] = dep[f] + 1;
    if(a[f] > a[u]){
        fa[u][0] = f;
    }
    else{
        int x = f;
        for(int i = 20;~i;i--){
            if(fa[x][i] && a[fa[x][i]] <= a[u]) x = fa[x][i];
        }
        fa[u][0] = fa[x][0];
    }
    for(int i = 1;i <= 20;i++) fa[u][i] = fa[fa[u][i - 1]][i - 1];
    for(int i = h[u];~i;i = ne[i]){
        int j = e[i];
        if(j == f) continue;
        dfs(j,u);
    }
}

int main() {
    memset(h,-1,sizeof h);
    read(n),read(m);    
    for(int i = 1;i <= n;i++) read(a[i]);
    for(int i = 1;i < n;i++){
        int x,y;
        read(x),read(y);
        add(x,y),add(y,x);
    }
    for(int i = 1;i <= m;i++){
        int u,v,val;
        read(u),read(v),read(val);
        a[n + i] = val;
        add(u,n + i),add(n + i,u);
        q[i] = v;
    }
    dfs(1,0);
    for(int j = 1;j <= m;j++){
        int ans = 0;
        int u = n + j,v = q[j];
        for(int i = 20;~i;i--){
            if(dep[fa[u][i]] >= dep[v]) ans += (1 << i),u = fa[u][i];
        }
        di(ans);
    }
}