标签:cas make put who track mem key mode dal
LCT:
分割、合并子树,路径上全部点的点权添加一个值,查询路径上点权的最大值
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 7HintWe 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
标签:cas make put who track mem key mode dal
原文地址:http://www.cnblogs.com/liguangsunls/p/7327884.html