Hdu 4010 Query on The Trees lct link-cut tree

Query on The Trees

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 2628    Accepted Submission(s): 1223

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

   
   
   
   
   
   
   
   


   
   
   
   

link-cut树模板练习中，就是比较慢不知道怎么优化

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define maxn 300007
#define inf  1000000000
struct Node{
    Node *fa,*ch[2];
    bool rev,root;
    int val,add,size;
    int max;
};
Node pool[maxn];
Node *nil,*tree[maxn];
int cnt = 0;
void init(){
    cnt = 1;
    nil = tree[0] = pool;
    nil->size = 0;
}
inline Node *newnode(int val,Node *f){
    pool[cnt].fa = f;
    pool[cnt].ch[0] = pool[cnt].ch[1] = nil;
    pool[cnt].rev = false;
    pool[cnt].root = true;
    pool[cnt].val = val;
    pool[cnt].add = 0;
    pool[cnt].size = 1;
    pool[cnt].max = val;
    return &pool[cnt++];
}
void update_rev(Node *x){
    if(x == nil) return ;
    x->rev = !x->rev;
    swap(x->ch[0],x->ch[1]);
}
//splay向上更新信息
void update(Node *x){
    x->size = x->ch[0]->size + x->ch[1]->size + 1;
    x->max=x->val;
    if(x->ch[0] != nil)
        if(x->max < x->ch[0]->max)
            x->max = x->ch[0]->max;
    if(x->ch[1] != nil)
        if(x->max < x->ch[1]->max)
            x->max = x->ch[1]->max;
}

void update_add(Node *x,int n){
    if(x == nil) return ;
    x->max += n ;
    x->val += n;
    x->add += n;
}
//splay下推信息
void pushdown(Node *x){
    if(x->add != 0){
        update_add(x->ch[0],x->add);
        update_add(x->ch[1],x->add);
        x->add = 0;
    }
    if(x->rev != false){
        update_rev(x->ch[0]);
        update_rev(x->ch[1]);
        x->rev = false;
    }
}
//splay在root-->x的路径下推信息
void push(Node *x){
    if(!x->root) push(x->fa);
    pushdown(x);
}
//将结点x旋转至splay中父亲的位置
void rotate(Node *x){
    Node *f = x->fa, *ff = f->fa;
    int t = (f->ch[1] == x);
    if(f->root) x->root = true, f->root = false;
    else ff->ch[ff->ch[1] == f] = x;
    x->fa = ff;
    f->ch[t] = x->ch[t^1];
    x->ch[t^1]->fa = f;
    x->ch[t^1] = f;
    f->fa = x;
    update(f);
}
//将结点x旋转至x所在splay的根位置
void splay(Node *x){
    push(x);
    Node *f, *ff;
    while(!x->root){
        f = x->fa,ff = f->fa;
        if(!f->root)
            if((ff->ch[1] == f) && (f->ch[1] == x)) rotate(f);
            else rotate(x);
        rotate(x);
    }
    update(x);
}
//将x到树根的路径并成一条path
Node *access(Node *x){
    Node *y = nil;
    while(x != nil){
        splay(x);
        x->ch[1]->root = true;
        (x->ch[1] = y)->root = false;
        update(x);
        y = x;
        x = x->fa;
    }
    return y;
}
//将结点x变成树根
void be_root(Node *x){
    access(x);
    splay(x);
    update_rev(x);
}
//将x连接到结点f上
void link(Node *x, Node *f){
    be_root(x);
    x->fa = f;
}
//将x,y分离
void cut(Node *x,Node *y){
    be_root(x);
    access(x);
    splay(y);
    y->fa = nil;
}
Node *find_root(Node *p){
    while(1){
        pushdown(p);
        if(p->ch[0] != nil) p = p->ch[0];
        else break;
    }
    return p;
}
Node *find_father(Node *p){
    pushdown(p);
    if(p->ch[0] == nil) return nil;
    p = p->ch[0];
    while(1){
        pushdown(p);
        if(p->ch[1] != nil) p = p->ch[1];
        else break;
    }
    return p;
}
struct Edge{
    int v,next;
};
Edge edge[2*maxn];
int head[maxn],ecnt;
int value[maxn];
void add_edge(int u,int v){
    edge[ecnt].v = v;
    edge[ecnt].next = head[u];
    head[u] = ecnt++;
    edge[ecnt].v = u;
    edge[ecnt].next = head[v];
    head[v] = ecnt++;
}
void dfs(int u,int f){
    tree[u] = newnode(value[u],tree[f]);
    for(int i = head[u]; i != -1;i = edge[i].next){
        if(edge[i].v == f) continue;
        dfs(edge[i].v,u);
    }
}
int main(){
    int n,m,u,v,w,t,x,y;
    Node *p;
    while(scanf("%d",&n)!=EOF){
        memset(head,-1,sizeof(head));
        ecnt = 0;
        init();
        for(int i = 0;i < n - 1; i++){
            scanf("%d%d",&u,&v);
            add_edge(u,v);
        }
        for(int i = 1;i <= n; i++)
            scanf("%d",&value[i]);
        dfs(1,0);
        scanf("%d",&m);
        while(m--){
            scanf("%d",&t);
            if(t == 1){
                scanf("%d%d",&x,&y);
                be_root(tree[x]);
                p = access(tree[y]);
                if(tree[x] == find_root(p)) printf("-1\n");
                else link(tree[x],tree[y]);
            }
            else if(t == 2){
                scanf("%d%d",&x,&y);
                be_root(tree[x]);
                p = access(tree[y]);
                if(tree[x] != find_root(p) || y == x) printf("-1\n");
                else {
                    splay(tree[y]);
                    p = find_father(tree[y]);
                    cut(tree[y],p);
                }
            }
            else if(t == 3){
                scanf("%d%d%d",&w,&x,&y);
                be_root(tree[x]);
                p = access(tree[y]);
                if(tree[x] != find_root(p)) printf("-1\n");
                else update_add(p,w);
            }
            else {
                scanf("%d%d",&x,&y);
                be_root(tree[x]);
                p = access(tree[y]);
                if(tree[x] != find_root(p))printf("-1\n");
                else printf("%d\n",p->max);
            }
        }
        printf("\n");
    }
    return 0;
}