hdu 4718 The LCIS on the Tree

The LCIS on the Tree

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 528    Accepted Submission(s): 165

Problem Description

For a sequence S₁, S₂, ... , S_N, and a pair of integers (i, j), if 1 <= i <= j <= N and S_i < S_i+1 < S_i+2 < ... < S_j-1 < S_j , then the sequence S_i, S_i+1, ... , S_j is a CIS(Continuous Increasing Subsequence). The longest CIS of a sequence is called the LCIS (Longest Continuous Increasing Subsequence).
Now we consider a tree rooted at node 1. Nodes have values. We have Q queries, each with two nodes u and v. You have to find the shortest path from u to v. And write down each nodes' value on the path, from u to v, inclusive. Then you will get a sequence, and please show us the length of its LCIS.

 

Input

The first line has a number T (T <= 10) , indicating the number of test cases.
For each test case, the first line is a number N (N <= 10⁵), the number of nodes in the tree.
The second line comes with N numbers v1, v2, v3 ... , v_N, describing the value of node 1 to node N. (1 <= v_i <= 10⁹)
The third line comes with N - 1 numbers p₂, p₃, p₄ ... , p_N, describing the father nodes of node 2 to node N. Node 1 is the root and will have no father.
Then comes a number Q, it is the number of queries. (Q <= 10⁵)
For next Q lines, each with two numbers u and v. As described above.

 

Output

For test case X, output "Case #X:" at the first line.
Then output Q lines, each with an answer to the query.
There should be a blank line *BETWEEN* each test case.

 

Sample Input

1

5

1 2 3 4 5

1 1 3 3

3

1 5

4 5

2 5

 

Sample Output

Case #1:

3

2

3

 

Source

2013 ACM/ICPC Asia Regional Online —— Warmup2

 

题意：给你一棵树，每个点都有一个权值，每个询问 u v 

求 u 到 v 路径上节点权值序列的最长上升子序列

 

思路：数链剖分，线段树维护左到右的最长上升子序列，和右到左的最长上升子序列，

合并注意点就好了。。

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
#include<set>
#include<stack>
#include<map>
#include<ctime>
#include<bitset>
#define LL long long
#define ll long long
#define INF 0x3f3f3f3f
#define maxn 100010
#define eps 1e-6
#define mod 1000000007
using namespace std;

struct node
{
    int ls,rs,lmax,rmax,max1,max2;
    int len,Lmax,Rmax;
    void init()
    {
        ls=rs=lmax=rmax=0;
        max1=max2=Lmax=Rmax=0;
        len=0;
    }
}tree[maxn*3];
int head[maxn],to[maxn*2],next1[maxn*2] ;
int tot,val[maxn],fa[maxn],top[maxn];
int dep[maxn],son[maxn],w[maxn];
int sz,ql,qr,n;
node v;

node Unit(node &a,node &b)
{
    node ans;
    ans.len=a.len+b.len ;
    ans.ls=a.ls ;
    ans.rs=b.rs ;
    ans.max1=max(a.max1,b.max1) ;
    ans.max2=max(a.max2,b.max2) ;

    if(a.rs < b.ls )
    {
        if(a.lmax==a.len)
           ans.lmax = a.lmax+b.lmax ;
        else ans.lmax=a.lmax;
        ans.max1=max(ans.max1,a.rmax+b.lmax) ;
        if(b.rmax==b.len)
           ans.rmax = a.rmax+b.rmax ;
        else ans.rmax=b.rmax;
    }
    else {
         ans.lmax = a.lmax ;
         ans.rmax = b.rmax ;
    }

    if(b.ls < a.rs )
    {
        if(a.Lmax==a.len)
          ans.Lmax = a.Lmax+b.Lmax ;
        else ans.Lmax=a.Lmax;
        ans.max2=max(ans.max2,a.Rmax+b.Lmax) ;
        if(b.Rmax == b.len)
          ans.Rmax = a.Rmax+b.Rmax ;
        else ans.Rmax=b.Rmax;
    }
    else {
         ans.Lmax = a.Lmax ;
         ans.Rmax = b.Rmax ;
    }
    return ans;
}
void Unit(int u,int v)
{
    to[tot]=v;next1[tot]=head[u];
    head[u]=tot++;
}
void dfs(int u,int f)
{
    son[u]=0;w[u]=1;
    int i,v;
    for( i = head[u] ; i != -1; i = next1[i])
    {
        v = to[i] ;
        if(v==f) continue ;
        fa[v]=u;
        dep[v]=dep[u]+1;
        dfs(v,u) ;
        if(w[v]>w[son[u]])
            son[u]=v;
        w[u] += w[v] ;
    }
}
void build(int u,int f,int t)
{
    w[u]=++sz ;
    top[u]=t;
    if(son[u])build(son[u],u,t) ;
    for( int i = head[u] ; i != -1 ; i = next1[i])
        if(son[u] != to[i]&&f != to[i])
         build(to[i],u,to[i]) ;
}
void init(int L,int R,int o)
{
    if(L==R)
    {
        tree[o].ls=tree[o].rs=son[L] ;
        tree[o].len=tree[o].max1=tree[o].max2=1;
        tree[o].lmax=tree[o].Lmax=1;
        tree[o].Rmax=tree[o].rmax = 1;
        return ;
    }
    int mid=(L+R)>>1;
    init(L,mid,o<<1) ;
    init(mid+1,R,o<<1|1) ;
    tree[o]=Unit(tree[o<<1],tree[o<<1|1]) ;
}

void find(int L,int R,int o)
{
    if(ql<=L&&qr>=R)
    {
        if(v.max1==0)v=tree[o] ;
        else v = Unit(v,tree[o]) ;
        return ;
    }
    int mid=(L+R)>>1 ;
    if(ql<=mid) find(L,mid,o<<1) ;
    if(qr>mid) find(mid+1,R,o<<1|1);
}

int solve(int u,int vv)
{
    int f1,f2;
    node v1,v2;
    v1.init();
    v2.init();
    f1=top[u];
    f2=top[vv];
    while(f1 != f2)
    {
        if(dep[f1]>dep[f2])
        {
            ql=w[f1];
            qr=w[u] ;
            v.init();
            find(1,sz,1) ;
            if(v1.max1==0)
                v1=v;
            else v1=Unit(v,v1) ;
            u=fa[f1];
            f1=top[u];
        }
        else{
            ql=w[f2];
            qr=w[vv] ;
            v.init();
            find(1,sz,1) ;
            if(v2.max1==0)
                v2=v;
            else v2=Unit(v,v2) ;
            vv=fa[f2] ;
            f2=top[vv] ;
        }
    }
    int ans=0;
    if(u==vv&&0){
        ans=max(v1.max2,v2.max1);
        if(v1.ls < v2.ls)
        {
            ans=max(ans,v1.Lmax+v2.lmax) ;
        }
        return ans;
    }
    else
    {
        if(dep[u]>dep[vv])
        {
            ql=w[vv];
            qr=w[u];
            v.init();
            find(1,sz,1) ;
            if(v1.max1==0)v1=v;
            else v1=Unit(v,v1) ;
        }
        else
        {
            ql=w[u];
            qr=w[vv];
            v.init();
            find(1,sz,1) ;
            if(v2.max1==0)v2=v;
            else v2=Unit(v,v2) ;
        }
    }
    ans=max(v1.max2,v2.max1);
    if(v1.ls < v2.ls)
    {
        ans=max(ans,v1.Lmax+v2.lmax) ;
    }
    return ans;
}

int main()
{
    int i,m,k,case1=0;
    int j ,T,x,y ;
   // freopen("in.txt","r",stdin);
   // freopen("yy.txt","w",stdout);
   bool hehe=false;
    cin >> T ;
    while(T--)
    {
        if(hehe)puts("") ;
        hehe=true;
        scanf("%d",&n) ;
        for( i =1 ; i <= n ;i++)
            scanf("%d",&val[i]) ;
        tot=0;
        memset(head,-1,sizeof(head)) ;
        for( i = 2 ; i <= n ;i++)
        {
            scanf("%d",&k) ;
            Unit(k,i) ;
            Unit(i,k) ;
        }
        dep[1]=0;
        fa[1]=1;
        sz=0;
        dfs(1,-1) ;
        build(1,-1,1) ;
        for( i = 1 ; i <= n ;i++)
        {
            son[w[i]]=val[i] ;
        }
        init(1,sz,1) ;
        scanf("%d",&m) ;
        printf("Case #%d:
",++case1);
        while(m--)
        {
            scanf("%d%d",&x,&y) ;
            printf("%d
",solve(x,y)) ;
        }
    }
    return 0 ;
}