HDU 4010 Query on The Trees

[align=left]Problem Description[/align]
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!

[align=left]Input[/align]
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.

[align=left]Output[/align]
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.

[align=left]Sample Input[/align]

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

[align=left]Sample Output[/align]

3
-1
7

[b]LCT模板[/b]

#include<cstdio>
#include<cstring>
#include<algorithm>
#define MN 300001
using namespace std;

int p,ca,f;
inline int read(){
p=0;ca=getchar();f=1;
while(ca<'0'||ca>'9') {if (ca=='-') f=-1;ca=getchar();}
while(ca>='0'&&ca<='9') p=p*10+ca-48,ca=getchar();
return p*f;
}
struct na{
int y,ne;
}b[MN*2];
int fa[MN],n,m,x,y,ch[MN][2],top,st[MN],key[MN],ma[MN],c[MN],l[MN],r[MN],num;
bool rev[MN];
inline int max(int a,int b){return a>b?a:b;}
inline bool isroot(int x){
return ch[fa[x]][0]!=x&&ch[fa[x]][1]!=x;
}
inline void pu(int x,int w){
if (!x) return;
c[x]+=w;key[x]+=w;ma[x]+=w;
}
inline void pd(int x){
if (c[x]){
pu(ch[x][0],c[x]);pu(ch[x][1],c[x]);
c[x]=0;
}
if (rev[x]){
rev[x]=0;rev[ch[x][0]]^=1;rev[ch[x][1]]^=1;
swap(ch[x][0],ch[x][1]);
}
}
inline void up(int x){
pd(x);pd(ch[x][0]);pd(ch[x][1]);
ma[x]=max(max(ma[ch[x][0]],ma[ch[x][1]]),key[x]);
}
inline void rot(int x){
int y=fa[x],kind=ch[y][1]==x;
if(!isroot(y)) ch[fa[y]][ch[fa[y]][1]==y]=x;
fa[x]=fa[y];
fa[y]=x;
ch[y][kind]=ch[x][!kind];
fa[ch[y][kind]]=y;
ch[x][!kind]=y;
up(y);up(x);
}
inline void splay(int x){
register int i;top=1;st[1]=x;
for (i=x;!isroot(i);i=fa[i]) st[++top]=fa[i];
for (i=top;i;i--) up(st[i]);
while(!isroot(x)){
if (isroot(fa[x])) rot(x);else
if ((ch[fa[fa[x]]][1]==fa[x])==(ch[fa[x]][1]==x)) rot(fa[x]),rot(x);else rot(x),rot(x);
}
}
inline void acc(int u){
int x=0;
while(u){
splay(u);
ch[u][1]=x;
u=fa[x=u];
}
}
inline int find(int x){
acc(x);splay(x);
while(ch[x][0]) x=ch[x][0];
return x;
}
inline bool qu(int x,int y){
if (find(x)==find(y)) return 1;else return 0;
}
inline void re(int x){
acc(x);splay(x);rev[x]^=1;
}
inline void in(int x,int y){
if (qu(x,y)){printf("-1\n");return;}
re(x);acc(y);
ch[y][1]=x;fa[x]=y;
}
inline void del(int x,int y){
if (!qu(x,y)||x==y){printf("-1\n");return;}
re(x);acc(y);splay(y);ch[y][0]=fa[ch[y][0]]=0;
}
inline void change(int x,int y,int w){
if (!qu(x,y)){printf("-1\n");return;}
re(x);acc(y);splay(y);pu(y,w);
}
inline int que(int x,int y){
if (!qu(x,y)) return -1;
re(x);acc(y);splay(y);
return ma[y];
}
inline void inl(int x,int y){
num++;
if (!l[x]) l[x]=num;else b[r[x]].ne=num;
b[num].y=y;b[num].ne=0;r[x]=num;
}
inline void dfs(int x){
for (int i=l[x];i;i=b[i].ne)
if (!fa[b[i].y]){
fa[b[i].y]=x;
dfs(b[i].y);
}
}
int o;
int main(){
register int i;
ma[0]=-1e9;
while(scanf("%d",&n)!=EOF){
num=0;
memset(fa,0,sizeof(fa));
memset(ch,0,sizeof(ch));
memset(c,0,sizeof(c));
memset(l,0,sizeof(l));
memset(rev,0,sizeof(rev));
for (i=1;i<n;i++){
x=read();y=read();
inl(x,y);
inl(y,x);
}
for(i=1;i<=n;i++) ma[i]=key[i]=read();
fa[1]=1;
dfs(1);
fa[1]=0;
m=read();
while(m--){
o=read();
int x,y,z;x=read();y=read();
if (o==1) in(x,y);else
if (o==2) del(x,y);else
if (o==3) z=read(),change(y,z,x);else
printf("%d\n",que(x,y));
}
printf("\n");
}
}

View Code