HDU 4010 Query on The Trees (动态树)

Query on The Trees

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 1527    Accepted Submission(s): 747

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

 

 

Recommend

lcy

 

 

 

动态树入门。

解释看代码：

/* ***********************************************
Author        :kuangbin
Created Time  :2013-9-4 0:13:15
File Name     :HDU4010.cpp
************************************************ */

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
//动态维护一组森林，要求支持一下操作:
//link(a,b) : 如果a,b不在同一颗子树中，则通过在a,b之间连边的方式，连接这两颗子树
//cut(a,b)  : 如果a,b在同一颗子树中，且a!=b,则将a视为这颗子树的根以后，切断b与其父亲结点的连接
//ADD(a,b,w): 如果a,b在同一颗子树中，则将a,b之间路径上所有点的点权增加w
//query(a,b): 如果a,b在同一颗子树中，返回a,b之间路径上点权的最大值
const int MAXN = 300010;
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);
}
//P函数先将根结点到r的路径上所有的结点的标记逐级下放
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;
}
//判断是否是同根(真实的树，非splay)
bool judge(int u,int v)
{
    while(pre[u]) u = pre[u];
    while(pre[v]) v = pre[v];
    return u == v;
}
//使r成为它所在的树的根
void mroot(int r)
{
    Access(r);
    Splay(r);
    Update_Rev(r);
}
//调用后u是原来u和v的lca,v和ch[u][1]分别存着lca的2个儿子
//(原来u和v所在的2颗子树)
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;
}
//使u成为u所在树的根，并且v和它父亲的边断开 
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 head[MAXN],tot;
void addedge(int u,int v)
{
    edge[tot].to = v;
    edge[tot].next = head[u];
    head[u] = tot++;
}
void dfs(int u)
{
    for(int i = head[u];i != -1; i = edge[i].next)
    {
        int v = edge[i].to;
        if(pre[v] != 0)continue;
        pre[v] = u;
        dfs(v);
    }
}

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int n,q,u,v;
    while(scanf("%d",&n) == 1)
    {
        tot = 0;
        for(int i = 0;i <= n;i++)
        {
            head[i] = -1;
            pre[i] = 0;
            ch[i][0] = ch[i][1] = 0;
            rev[i] = 0;
            add[i] = 0;
            rt[i] = true;
        }
        Max[0] = -2000000000;
        for(int i = 1;i < n;i++)
        {
            scanf("%d%d",&u,&v);
            addedge(u,v);
            addedge(v,u);
        }
        for(int i = 1;i <= n;i++)
        {
            scanf("%d",&key[i]);
            Max[i] = key[i];
        }
        scanf("%d",&q);
        pre[1] = -1;
        dfs(1);
        pre[1] = 0;
        int op;
        while(q--)
        {
            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 w,x,y;
                scanf("%d%d%d",&w,&x,&y);
                ADD(x,y,w);
            }
            else
            {
                int x,y;
                scanf("%d%d",&x,&y);
                query(x,y);
            }
        }
        printf("\n");
    }
    return 0;
}