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LCT:

切割、合并子树,路径上所有点的点权增加一个值,查询路径上点权的最大值

Query on The Trees

5
1 2
2 4
2 5
1 3
1 2 3 4 5
6
4 2 3
2 1 2
4 2 3
1 3 5
3 2 1 4
4 1 4

3
-1
7

Hint
We define the illegal situation of different operations: 
In first operation: if node x and y belong to a same tree, we think it‘s illegal. 
In second operation: if x = y or x and y not belong to a same tree, we think it‘s illegal. 
In third operation: if x and y not belong to a same tree, we think it‘s illegal. 
In fourth operation: if x and y not belong to a same tree, we think it‘s illegal. 
 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int maxn=330000;

int ch[maxn][2],pre[maxn],key[maxn];
int add[maxn],rev[maxn],Max[maxn];
bool rt[maxn];

void update_add(int r,int d)
{
  if(!r) return ;
  key[r]+=d;
  add[r]+=d;
  Max[r]+=d;
}

void update_rev(int r)
{
  if(!r) return ;
  swap(ch[r][0],ch[r][1]);
  rev[r]^=1;
}

void push_down(int r)
{
  if(add[r])
    {
      update_add(ch[r][0],add[r]);
      update_add(ch[r][1],add[r]);
      add[r]=0;
    }
  if(rev[r])
    {
      update_rev(ch[r][0]);
      update_rev(ch[r][1]);
      rev[r]=0;
    }
}

void push_up(int r)
{
  Max[r]=max(max(Max[ch[r][0]],Max[ch[r][1]]),key[r]);
}

void Rotate(int x)
{
  int y=pre[x],kind=(ch[y][1]==x);
  ch[y][kind]=ch[x][!kind];
  pre[ch[y][kind]]=y;
  pre[x]=pre[y];
  pre[y]=x;
  ch[x][!kind]=y;
  if(rt[y]) rt[y]=false,rt[x]=true;
  else ch[pre[x]][ch[pre[x]][1]==y]=x;
  push_up(y);
}

void P(int r)
{
  if(!rt[r]) P(pre[r]);
  push_down(r);
}

void Splay(int r)
{
  P(r);
  while(!rt[r])
    {
      int f=pre[r],ff=pre[f];
      if(rt[f]) Rotate(r);
      else if((ch[ff][1]==f)==(ch[f][1]==r)) Rotate(f),Rotate(r);
      else Rotate(r),Rotate(r);
    }
  push_up(r);
}

int Access(int x)
{
  int y=0;
  for(;x;x=pre[y=x])
    {
      Splay(x);
      rt[ch[x][1]]=true; rt[ch[x][1]=y]=false;
      push_up(x);
    }
  return y;
}

bool judge(int u,int v)
{
  while(pre[u]) u=pre[u];
  while(pre[v]) v=pre[v];
  return u==v;
}

void mroot(int r)
{
  Access(r);
  Splay(r);
  update_rev(r);
}

void lca(int &u,int &v)
{
  Access(v); v=0;
  while(u)
    {
      Splay(u);
      if(!pre[u]) return ;
      rt[ch[u][1]]=true;
      rt[ch[u][1]=v]=false;
      push_up(u);
      u=pre[v=u];
    }
}

void link(int u,int v)
{
  if(judge(u,v))
    {
      puts("-1");
      return ;
    }
  mroot(u);
  pre[u]=v;
}

void cut(int u,int v)
{
  if(u==v||!judge(u,v))
    {
      puts("-1");
      return ;
    }
  mroot(u);
  Splay(v);
  pre[ch[v][0]]=pre[v];
  pre[v]=0;
  rt[ch[v][0]]=true;
  ch[v][0]=0;
  push_up(v);
}

void Add(int u,int v,int w)
{
  if(!judge(u,v))
    {
      puts("-1"); return ;
    }
  lca(u,v);
  update_add(ch[u][1],w);
  update_add(v,w);
  key[u]+=w;
  push_up(u);
}

void query(int u,int v)
{
  if(!judge(u,v))
    {
      puts("-1");
      return ;
    }
  lca(u,v);
  printf("%d\n",max(max(Max[v],Max[ch[u][1]]),key[u]));
}

struct Edge
{
  int to,next;
}edge[maxn*2];

int Adj[maxn],Size=0;

void init()
{
  memset(Adj,-1,sizeof(Adj)); Size=0;
}

void add_edge(int u,int v)
{
  edge[Size].to=v;
  edge[Size].next=Adj[u];
  Adj[u]=Size++;
}

void dfs(int u)
{
  for(int i=Adj[u];~i;i=edge[i].next)
    {
      int v=edge[i].to;
      if(pre[v]!=0) continue;
      pre[v]=u;
      dfs(v);
    }
}
int n;

int main()
{
  while(scanf("%d",&n)!=EOF)
    {
      init();
      for(int i=0;i<n+10;i++)
        {
          pre[i]=0; ch[i][0]=ch[i][1]=0;
          rev[i]=0; add[i]=0; rt[i]=true;
        }

      for(int i=0;i<n-1;i++)
        {
          int u,v;
          scanf("%d%d",&u,&v);
          add_edge(u,v);
          add_edge(v,u);
        }
      pre[1]=-1; dfs(1); pre[1]=0;

      for(int i=1;i<=n;i++)
        {
          scanf("%d",key+i);
          Max[i]=key[i];
        }

      int q;
      scanf("%d",&q);
      while(q--)
        {
          int op;
          scanf("%d",&op);
          if(op==1)
            {
              int x,y;
              scanf("%d%d",&x,&y);
              link(x,y);
            }
          else if(op==2)
            {
              int x,y;
              scanf("%d%d",&x,&y);
              cut(x,y);
            }
          else if(op==3)
            {
              int x,y,w;
              scanf("%d%d%d",&w,&x,&y);
              Add(x,y,w);
            }
          else if(op==4)
            {
              int x,y;
              scanf("%d%d",&x,&y);
              query(x,y);
            }
        }
      putchar(10);
    }
  return 0;
}

HDOJ 4010 Query on The Trees LCT

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原文地址：http://blog.csdn.net/ck_boss/article/details/40481385