Query on The Trees
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 2599 Accepted Submission(s): 1213
Problem Description
We have met so many problems on the tree, so today we will have a query problem on a set of trees.
There are N nodes, each node will have a unique weight Wi. We will have four kinds of operations on it and you should solve them efficiently. Wish you have fun!
Input
There are multiple test cases in our dataset.
For each case, the first line contains only one integer N.(1 ≤ N ≤ 300000) The next N‐1 lines each contains two integers x, y which means there is an edge between them. It also means we will give you one tree initially.
The next line will contains N integers which means the weight Wi of each node. (0 ≤ Wi ≤ 3000)
The next line will contains an integer Q. (1 ≤ Q ≤ 300000) The next Q lines will start with an integer 1, 2, 3 or 4 means the kind of this operation.
1. Given two integer x, y, you should make a new edge between these two node x and y. So after this operation, two trees will be connected to a new one.
2. Given two integer x, y, you should find the tree in the tree set who contain node x, and you should make the node x be the root of this tree, and then you should cut the edge between node y and its parent. So after this operation, a tree will be separate
into two parts.
3. Given three integer w, x, y, for the x, y and all nodes between the path from x to y, you should increase their weight by w.
4. Given two integer x, y, you should check the node weights on the path between x and y, and you should output the maximum weight on it.
Output
For each query you should output the correct answer of it. If you find this query is an illegal operation, you should output ‐1.
You should output a blank line after each test case.
Sample Input
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
Sample Output
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.
Source
The 36th ACM/ICPC Asia Regional Dalian Site —— Online
Contest
代码：
/* ***********************************************
Author :rabbit
Created Time :2014/11/1 10:28:39
File Name :4.cpp
************************************************ */
#pragma comment(linker, "/STACK:102400000,102400000")
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <string>
#include <time.h>
#include <math.h>
#include <queue>
#include <stack>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define eps 1e-8
#define pi acos(-1.0)
typedef long long ll;
const int maxn=300300;
struct Node ;
Node *null;
struct Node{
Node *fa,*ch[2];
int Max,val,rev,add;
inline void clear(int _val){
fa=ch[0]=ch[1]=null;
val=Max=_val;
rev=0;
add=0;
}
inline void push_up(){
Max=max(val,max(ch[0]->Max,ch[1]->Max));
}
inline void setc(Node *p,int d){
ch[d]=p;
p->fa=this;
}
inline bool d(){
return fa->ch[1]==this;
}
inline bool isroot(){
return fa==null||fa->ch[0]!=this&&fa->ch[1]!=this;
}
inline void flip(){
if(this==null)return;
swap(ch[0],ch[1]);
rev^=1;
}
inline void update_add(int w){
if(this==null)return;
val+=w;
add+=w;
Max+=w;
}
inline void push_down(){
if(rev){
ch[0]->flip();ch[1]->flip();rev=0;
}
if(add){
ch[0]->update_add(add);
ch[1]->update_add(add);
add=0;
}
}
inline void go(){
if(!isroot())fa->go();
push_down();
}
inline void rot(){
Node *f=fa,*ff=fa->fa;
int c=d(),cc=fa->d();
f->setc(ch[!c],c);
this->setc(f,!c);
if(ff->ch[cc]==f)ff->setc(this,cc);
else this->fa=ff;
f->push_up();
}
inline Node *splay(){
go();
while(!isroot()){
if(!fa->isroot())d()==fa->d()?fa->rot():rot();
rot();
}
push_up();
return this;
}
inline Node *access(){
for(Node *p=this,*q=null;p!=null;q=p,p=p->fa){
p->splay()->setc(q,1);
p->push_up();
}
return splay();
}
inline Node *find_root(){
Node *x;
for(x=access();x->push_down(),x->ch[0]!=null;x=x->ch[0]);
return x;
}
void make_root(){
access()->flip();
}
void cut(){
access();
ch[0]->fa=null;
ch[0]=null;
push_up();
}
void cut(Node *x){
if(this==x||find_root()!=x->find_root())puts("-1");
else{
x->make_root();
cut();
}
}
void link(Node *x){
if(find_root()==x->find_root())puts("-1");
else{
make_root();fa=x;
}
}
};
Node pool[maxn],*tail,*node[maxn];
struct Edge{
int next,to;
}edge[2*maxn];
int head[maxn],tot;
inline void addedge(int u,int v){
edge[tot].to=v;
edge[tot].next=head[u];
head[u]=tot++;
}
void dfs(int u,int pre){
for(int i=head[u];i!=-1;i=edge[i].next){
int v=edge[i].to;
if(v==pre)continue;
node[v]->fa=node[u];
dfs(v,u);
}
}
void ADD(Node *x,Node *y,int w){
x->access();
for(x=null;y!=null;x=y,y=y->fa){
y->splay();
if(y->fa==null){
y->ch[1]->update_add(w);
x->update_add(w);
y->val+=w;
y->push_up();
return;
}
y->setc(x,1);
y->push_up();
}
}
int ask(Node *x,Node *y){
x->access();
for(x=null;y!=null;x=y,y=y->fa){
y->splay();
if(y->fa==null)
return max(y->val,max(y->ch[1]->Max,x->Max));
y->setc(x,1);
y->push_up();
}
}
int main()
{
//freopen("data.in","r",stdin);
//freopen("data.out","w",stdout);
int n;
while(~scanf("%d",&n)){
for(int i=1;i<=n;i++)head[i]=-1;tot=0;
for(int i=1;i<n;i++){
int u,v;
scanf("%d%d",&u,&v);
addedge(u,v);
addedge(v,u);
}
tail=pool;
null=tail++;
null->clear(-INF);
for(int i=1;i<=n;i++){
node[i]=tail++;
int v;scanf("%d",&v);
node[i]->clear(v);
}
dfs(1,1);
int m,op,x,y,w;
scanf("%d",&m);
while(m--){
scanf("%d",&op);
if(op==1){
scanf("%d%d",&x,&y);
node[x]->link(node[y]);
}
else if(op==2){
scanf("%d%d",&x,&y);
node[y]->cut(node[x]);
}
else if(op==3){
scanf("%d%d%d",&w,&x,&y);
if(node[x]->find_root()!=node[y]->find_root())puts("-1");
else ADD(node[x],node[y],w);
}
else{
scanf("%d%d",&x,&y);
if(node[x]->find_root()!=node[y]->find_root())puts("-1");
else printf("%d\n",ask(node[x],node[y]));
}
}
puts("");
}
return 0;
}