[HDU4010]Query on The Trees-动态树LCT(Link Cut Tree)

最新推荐文章于 2019-09-19 01:14:54 发布

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Query on The Trees

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.

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.
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.
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.
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.

感觉咱快忘记LCT怎么打了……于是决定找一道模板题练一练……
然后咱就在连样例都过不去的状态下调了一个多小时(╯‵□′)╯︵┻━┻~
总算过了样例一交，居然还Presentation Error……
咱的心情可以用下面的一个表情形容：

(╯‵□′)╯︵┻━┻

思路:
模板题能有什么思路……
除了3操作外都是LCT必备的操作啊……2操作还很贴心地把Cut操作的过程口述了一遍……
3操作直接makeroot其中一个点，旋转另一个节点并给它打tag就好了~
模板题就是模板题~(然而连样例都过不去的你有什么资格说啊喂)

P.S:千万不要学习咱这长到了一种新高度的代码排版风格……

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>

using namespace std;

const int N=300233;

inline int maxx(int a,int b){if(a>b)return a;return b;}

inline int read()
{
    int x=0;
    char ch=getchar();
    while(ch<'0' || '9'<ch)ch=getchar();
    while('0'<=ch && ch<='9')
    {
        x=(x<<1)+(x<<3)+(ch^48);
        ch=getchar();
    }
    return x;
}

int n;

int to[N],nxt[N],beg[N],tot;

void add(int u,int v)
{
    to[++tot]=v;
    nxt[tot]=beg[u];
    beg[u]=tot;
    to[++tot]=u;
    nxt[tot]=beg[v];
    beg[v]=tot;
}

struct LCT
{
    int fa[N],ch[N][2],rev[N];
    int mx[N],tag[N],w[N];
    int q[N],top;

    void init()
    {
        memset(tag,0,sizeof(tag));
        memset(rev,0,sizeof(rev));
        memset(fa,0,sizeof(fa));
        memset(ch,0,sizeof(ch));
        memset(mx,0,sizeof(mx));
        memset(w,0,sizeof(w));
        top=0;
    }

    void read(int val)
    {
        w[++top]=val;
    }

    bool isroot(int x)
    {
        return (ch[fa[x]][0]!=x) && (ch[fa[x]][1]!=x);
    }

    void push_up(int x)
    {
        mx[x]=maxx(w[x],maxx(mx[ch[x][0]],mx[ch[x][1]]));
    }

    void push_down(int x)
    {
        int l=ch[x][0],r=ch[x][1];
        if(rev[x])
        {
            swap(ch[x][0],ch[x][1]);
            if(l)rev[l]^=1;
            if(r)rev[r]^=1;
            rev[x]^=1;
        }
        if(tag[x])
        {
            if(l)w[l]+=tag[x],mx[l]+=tag[x],tag[l]+=tag[x];
            if(r)w[r]+=tag[x],mx[r]+=tag[x],tag[r]+=tag[x];
            tag[x]=0;
        }
    }

    void rotate(int x)
    {
        int faa=fa[x],gra=fa[faa];
        int l=(ch[faa][0]!=x),r=l^1;

        if(!isroot(faa))
        {
            if(ch[gra][0]==faa)ch[gra][0]=x;
            else ch[gra][1]=x;
        }
        fa[x]=gra;fa[faa]=x;
        fa[ch[x][r]]=faa;
        ch[faa][l]=ch[x][r];
        ch[x][r]=faa;
        push_up(faa);push_up(x);
    }

    void splay(int x)
    {
        q[top=1]=x;
        for(int i=x;!isroot(i);i=fa[i])
            q[++top]=fa[i];
        while(top)push_down(q[top--]);

        while(!isroot(x))
        {
            int faa=fa[x],gra=fa[faa];
            if(!isroot(faa))
            {
                if((ch[gra][0]==faa)^(ch[faa][0]==x))rotate(x);
                else rotate(faa);
            }
            rotate(x);
        }
    }

    void access(int x)
    {
        for(int t=0;x;t=x,x=fa[x])
        {
            splay(x);
            ch[x][1]=t;
            push_up(x);
        }
    }

    void makeroot(int x)
    {
        access(x);splay(x);rev[x]^=1;
    }

    void link(int x,int y)
    {
        makeroot(x);fa[x]=y;
    }

    void cut(int x,int y)
    {
        makeroot(x);
        access(y);splay(y);
        ch[y][0]=fa[ch[y][0]]=0;
        push_up(y);
    }

    int find(int x)
    {
        access(x);
        splay(x);
        while(ch[x][0])x=ch[x][0];
        return x;
    }

    void add(int x,int y,int val)
    {
        makeroot(x);
        access(y);splay(y);
        tag[y]+=val;
        mx[y]+=val;
        w[y]+=val;
    }

    int query(int x,int y)
    {
        makeroot(x);
        access(y);
        splay(y);
        return mx[y];
    }
}koishi;

int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        koishi.init();
        memset(beg,0,sizeof(beg));
        tot=0;

        for(int i=1;i<n;i++)
            add(read(),read());

        for(int i=1;i<=n;i++)
            koishi.read(read());

        koishi.q[koishi.top=1]=1;
        for(int j=1;j<=koishi.top;j++)
        {
            int now=koishi.q[j];
            for(int i=beg[now],v;i;i=nxt[i])
                if((v=to[i])!=koishi.fa[now])
                {
                    koishi.fa[v]=now;
                    koishi.q[++koishi.top]=v;
                }
        }

        int q=read();
        while(q--)
        {
            int type=read(),x=read(),y=read(),w;
            switch(type)
            {
                case 1:
                    if(koishi.find(x)==koishi.find(y))puts("-1");
                    else koishi.link(x,y);
                break;

                case 2:
                    if(x==y || koishi.find(x)!=koishi.find(y))puts("-1");
                    else koishi.cut(x,y);
                break;

                case 3:
                    if(koishi.find(y)!=koishi.find(w=read()))puts("-1");
                    else koishi.add(y,w,x);
                break;

                case 4:
                    if(koishi.find(x)!=koishi.find(y))puts("-1");
                    else printf("%d\n",koishi.query(x,y));
            }
        }
        puts("");
    }

    return 0;
}