牛客——选点(dfs序+LIS)
牛客——选点(dfs序+LIS)
思路:
根据题意,选出来的子树应该满足:左子树的点的权值>右子树的点的权值>根节点的权值。
对树按照根右左的顺序求dfs序,在转化后的序列里求最长上升子序列即可。
代码:
///#pragma GCC optimize(3)
///#pragma GCC optimize("Ofast","unroll-loops","omit-frame-pointer","inline")
///#pragma GCC optimize(2)
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll,ll>PLL;
typedef pair<int,int>PII;
typedef pair<double,double>PDD;
#define I_int ll
inline ll read()
{
ll x=0,f=1;
char ch=getchar();
while(ch<'0'||ch>'9')
{
if(ch=='-')f=-1;
ch=getchar();
}
while(ch>='0'&&ch<='9')
{
x=x*10+ch-'0';
ch=getchar();
}
return x*f;
}
char F[200];
inline void out(I_int x)
{
if (x == 0) return (void) (putchar('0'));
I_int tmp = x > 0 ? x : -x;
if (x < 0) putchar('-');
int cnt = 0;
while (tmp > 0)
{
F[cnt++] = tmp % 10 + '0';
tmp /= 10;
}
while (cnt > 0) putchar(F[--cnt]);
//cout<<" ";
}
ll ksm(ll a,ll b,ll p)
{
ll res=1;
while(b)
{
if(b&1)res=res*a%p;
a=a*a%p;
b>>=1;
}
return res;
}
const int inf=0x3f3f3f3f,mod=998244353;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const int maxn=1e5+100,maxm=3e5+7,N=1e6+7;
const double PI = atan(1.0)*4;
int w[maxn],n;
int b[maxn],timetmp;
PII e[maxn];
int q[maxn],cnt;
int Find(int x){
int l=1,r=cnt;
while(l<r){
int mid=(l+r)>>1;
if(q[mid]>=x) r=mid;
else l=mid+1;
}
return l;
}
void dfs(int u){
b[++timetmp]=w[u];
if(e[u].second) dfs(e[u].second);///注意顺序
if(e[u].first) dfs(e[u].first);
}
int main(){
///输入
n=read();
for(int i=1;i<=n;i++) w[i]=read();
for(int i=1;i<=n;i++){
e[i].first=read(),e[i].second=read();
}
///求根右左的dfs序
dfs(1);
///nlogn求LIS
q[++cnt]=b[1];
for(int i=2;i<=timetmp;i++)
if(b[i]>q[cnt]) q[++cnt]=b[i];
else{
int tmp=Find(b[i]);
q[tmp]=b[i];
}
cout<<cnt<<endl;
return 0;
}