Atcoder - 165F Lis on Tree

传送门：Atcoder - 165F

思路分析

回溯的时候删除该点的贡献就行了，然后就是用线段树优化lis的过程了

样例输入

10
1 2 5 3 4 6 7 3 2 4
1 2
2 3
3 4
4 5
3 6
6 7
1 8
8 9
9 10

样例输出

1
2
3
3
4
4
5
2
2
3

AC代码

#include <bits/stdc++.h>
#define ls x<<1
#define rs x<<1|1
#define fi first
#define se second
#define ll long long
#define pb push_back
#define mp make_pair
#define fun function
#define sz(x) (x).size()
#define lowbit(x) (x)&(-x)
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))

namespace FastIO {
#define BUF_SIZE 100000
#define OUT_SIZE 100000
    bool IOerror=0;
    inline char nc() {
        static char buf[BUF_SIZE],*p1=buf+BUF_SIZE,*pend=buf+BUF_SIZE;
        if(p1==pend) {
            p1=buf;
            pend=buf+fread(buf,1,BUF_SIZE,stdin);
            if(pend==p1) {
                IOerror=1;
                return -1;
            }
        }
        return *p1++;
    }
    inline bool blank(char ch) {
        return ch==' '||ch=='\n'||ch=='\r'||ch=='\t';
    }
    template<class T> inline bool read(T &x) {
        bool sign=0;
        char ch=nc();
        x=0;
        for(; blank(ch); ch=nc());
        if(IOerror)return false;
        if(ch=='-')sign=1,ch=nc();
        for(; ch>='0'&&ch<='9'; ch=nc())x=x*10+ch-'0';
        if(sign)x=-x;
        return true;
    }
    template<class T,class... U>bool read(T& h,U&... t) {
        return read(h)&&read(t...);
    }
#undef OUT_SIZE
#undef BUF_SIZE
};
using namespace std;
using namespace FastIO;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const int INF = 0x3f3f3f3f;
const int N = 2e5+10;

struct node {
    int l,r,mid,ma;
} tree[N<<2];

int a[N],ans[N];
vector<int>e[N];
int n;

void pp(int x) {
    tree[x].ma=max(tree[ls].ma,tree[rs].ma);
}

void build(int x,int l,int r) {
    tree[x]= {l,r,l+r>>1,0};
    if(l==r) {
        return ;
    }
    int mid=tree[x].mid;
    build(ls,l,mid);
    build(rs,mid+1,r);
}

void modify(int x,int pos,int k) {
    if(tree[x].l==pos && tree[x].r==pos) {
        tree[x].ma=k;
        return ;
    }
    int mid=tree[x].mid;
    if(pos<=mid) modify(ls,pos,k);
    if(pos>mid) modify(rs,pos,k);
    pp(x);
}

int query(int x,int l,int r) {
    if(l<=tree[x].l&&tree[x].r<=r) {
        return tree[x].ma;
    }
    int mid=tree[x].mid;
    int res=0;
    if(l<=mid) res=max(res,query(ls,l,r));
    if(r>mid) res=max(res,query(rs,l,r));
    return res;
}

void dfs(int u,int fa) {
    int now=a[u];
    int sum=query(1,1,now-1)+1;
    int pre=query(1,now,now);
    modify(1,now,max(sum,pre));
    ans[u]=query(1,1,n+5);
    for(auto v:e[u]) {
        if(v==fa) continue;
        dfs(v,u);
    }
    modify(1,now,pre);
}

int main() {

#ifdef xiaofan
    freopen("1.in","r",stdin);
    freopen("1.out","w",stdout);
#endif

    read(n);
    vector<int>b;
    for(int i=1; i<=n; i++) {
        read(a[i]);
        b.pb(a[i]);
    }
    sort(all(b));
    b.erase(unique(all(b)),b.end());
    for(int i=1; i<=n; i++) a[i]=lower_bound(all(b),a[i])-b.begin()+2;
    for(int i=1; i<n; i++) {
        int u,v;
        read(u,v);
        e[u].pb(v);
        e[v].pb(u);
    }
    build(1,1,n+5);
    dfs(1,1);
    for(int i=1; i<=n; i++)    cout<<ans[i]<<"\n";





    return 0;
}