HDU 4010 Query on The Trees (動態樹)

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

 

 

 

動態樹入門。

解釋看代碼：

/* ***********************************************
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;
//動態維護一組森林，要求支持一下操作:
//link(a,b) : 如果a,b不在同一顆子樹中，則通過在a,b之間連邊的方式，連接這兩顆子樹
//cut(a,b)  : 如果a,b在同一顆子樹中，且a!=b,則將a視為這顆子樹的根以后，切斷b與其父親結點的連接
//ADD(a,b,w): 如果a,b在同一顆子樹中，則將a,b之間路徑上所有點的點權增加w
//query(a,b): 如果a,b在同一顆子樹中，返回a,b之間路徑上點權的最大值
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函數先將根結點到r的路徑上所有的結點的標記逐級下放
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;
}
//判斷是否是同根(真實的樹，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成為它所在的樹的根
void mroot(int r)
{
    Access(r);
    Splay(r);
    Update_Rev(r);
}
//調用后u是原來u和v的lca,v和ch[u][1]分別存着lca的2個兒子
//(原來u和v所在的2顆子樹)
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成為u所在樹的根，並且v和它父親的邊斷開 
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;
}