H - The LCIS on the Tree HDU - 4718

The LCIS on the Tree

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 1615    Accepted Submission(s): 457

题目链接

https://vjudge.net/problem/UVA-12655

 

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

题意

在一棵树上，每次求一条路径最长连续上升子序列。

题解

一开始看错题，以为求最长上升子序列，然后想了好久毫无思路，一看题解发现原来我看错题了。

连续的话就不难了。

树链剖分，然后线段树记录一下该区间最长上升子序列，左端点开始的最长上升子序列长度，右端点。。。

反正就是考虑如果知道左子树和右子树的信息，如何合并。显然合并就是中间可能延长，其他的都一样。

但是在树上，x-->lca(x,y)　和　lca(x,y)-->是不一样的，x-->lca(x,y)需要求最长下降子序列，然后就是一对细节需要处理，具体看代码吧。

代码

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define N 500050
struct Tree{int l,r,lb,rb,lI,rI,lD,rD,Imx,Dmx;}tr[N<<2];
struct Edge{int from,to,s;}edges[N<<1];
int n,m,cas,w[N];
int tot,last[N];
int cnt,fa[N],dp[N],size[N],son[N],rk[N],kth[N],top[N];
template<typename T>void read(T&x)
{
  ll k=0; char c=getchar();
  x=0;
  while(!isdigit(c)&&c!=EOF)k^=c=='-',c=getchar();
  if (c==EOF)exit(0);
  while(isdigit(c))x=x*10+c-'0',c=getchar();
  x=k?-x:x;
}
void read_char(char &c)
{while(!isalpha(c=getchar())&&c!=EOF);}
void AddEdge(int x,int y)
{
  edges[++tot]=Edge{x,y,last[x]};
  last[x]=tot;
}
void dfs1(int x,int pre)
{
  fa[x]=pre;
  dp[x]=dp[pre]+1;
  size[x]=1;
  son[x]=0;
  for(int i=last[x];i;i=edges[i].s)
    {
      Edge &e=edges[i];
      if (e.to==pre)continue;
      dfs1(e.to,x);
      size[x]+=size[e.to];
      if(size[e.to]>size[son[x]])son[x]=e.to;
    }
}
void dfs2(int x,int y)
{
  rk[x]=++cnt;
  kth[cnt]=x;
  top[x]=y;
  if (son[x]==0)return;
  dfs2(son[x],y);
  for(int i=last[x];i;i=edges[i].s)
    {
      Edge &e=edges[i];
      if (e.to==fa[x]||e.to==son[x])continue;
      dfs2(e.to,e.to);
    }
}
void init(Tree &a){a=Tree{0,0,0,0,0,0,0,0,0,0};}
template<typename T>void he(T &c,T a, T b)
{
  if (a.Imx==0){c=b;return;}
  if (b.Imx==0){c=a;return;}
  c.l=a.l; c.r=b.r;
  int lena=a.r-a.l+1,lenb=b.r-b.l+1;
  c.lb=a.lb; c.rb=b.rb;
  c.lI=lena==a.lI&&a.rb<b.lb?lena+b.lI:a.lI;
  c.lD=lena==a.lD&&a.rb>b.lb?lena+b.lD:a.lD;
  c.rI=lenb==b.rI&&a.rb<b.lb?lenb+a.rI:b.rI;
  c.rD=lenb==b.rD&&a.rb>b.lb?lenb+a.rD:b.rD;
  c.Imx=max(a.Imx,b.Imx);
  c.Imx=max(c.Imx,a.rb<b.lb?a.rI+b.lI:0);
  c.Dmx=max(a.Dmx,b.Dmx);
  c.Dmx=max(c.Dmx,a.rb>b.lb?a.rD+b.lD:0);
}
void bt(int x,int l,int r)
{
  tr[x].l=l; tr[x].r=r;
  if (l==r)
    {
      tr[x]={l,r,w[kth[l]],w[kth[l]],1,1,1,1,1,1};
      return;
    }
  int mid=(l+r)>>1;
  bt(x<<1,l,mid);
  bt(x<<1|1,mid+1,r);
  he(tr[x],tr[x<<1],tr[x<<1|1]);
}
Tree query(int x,int l,int r)
{
  if (l<=tr[x].l&&tr[x].r<=r)
    return tr[x];
  int mid=(tr[x].l+tr[x].r)>>1;
  Tree tp1,tp2,tp; init(tp1); init(tp2);
  if (l<=mid)tp1=query(x<<1,l,r);
  if (mid<r)tp2=query(x<<1|1,l,r);
  he(tp,tp1,tp2);
  return tp;
}
int get_max(int x,int y)
{
  int fx=top[x],fy=top[y],ans=0;
  Tree tpx,tpy,tp;
  init(tpx); init(tpy);
  while(fx!=fy)
    {
      if (dp[fx]>dp[fy])
    {
      tp=query(1,rk[fx],rk[x]);
      he(tpx,tp,tpx);
      x=fa[fx]; fx=top[x];
    }
      else
    {
      tp=query(1,rk[fy],rk[y]);
      he(tpy,tp,tpy);
      y=fa[fy]; fy=top[y];
    }
    }
  if (dp[x]>dp[y])
    {
      tp=query(1,rk[y],rk[x]);
      he(tpx,tp,tpx);
    }
  else
    {
      tp=query(1,rk[x],rk[y]);
      he(tpy,tp,tpy);
    }
  ans=max(tpx.Dmx,tpy.Imx);
  ans=max(ans,tpx.lb<tpy.lb?tpx.lD+tpy.lI:0);
  return ans;
}
void work()
{
  if (cas)printf("\n");
  printf("Case #%d:\n",++cas);
  read(n);
  for(int i=1;i<=n;i++)read(w[i]);
  for(int i=2;i<=n;i++)
    {
      int x;
      read(x);
      AddEdge(x,i);
    }
  dfs1(1,0);
  dfs2(1,1);
  bt(1,1,n);
  read(m);
  for(int i=1;i<=m;i++)
    {
      int x,y;
      read(x); read(y);
      printf("%d\n",get_max(x,y));
    }
}
void clear()
{
  cnt=0; tot=0;
  memset(last,0,sizeof(last));
}
int main()
{
#ifndef ONLINE_JUDGE
  freopen("aa.in","r",stdin);
#endif
  int q;
  read(q);
  while(q--)
    {
      clear();
      work();
    } 
}