spoj 375 QTREE - Query on a tree 树链剖分 LCT 动态树

QTREE - Query on a tree

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You are given a tree (an acyclic undirected connected graph) with N nodes, and edges numbered 1, 2, 3...N-1.

We will ask you to perfrom some instructions of the following form:

• CHANGE i ti : change the cost of the i-th edge to ti
  or
• QUERY a b : ask for the maximum edge cost on the path from node a to node b

Input

The first line of input contains an integer t, the number of test cases (t <= 20). t test cases follow.

For each test case:

• In the first line there is an integer N (N <= 10000),
• In the next N-1 lines, the i-th line describes the i-th edge: a line with three integers a b c denotes an edge between a, b of cost c (c <= 1000000),
• The next lines contain instructions "CHANGE i ti" or "QUERY a b",
• The end of each test case is signified by the string "DONE".

There is one blank line between successive tests.

Output

For each "QUERY" operation, write one integer representing its result.

Example

Input:
1

3
1 2 1
2 3 2
QUERY 1 2
CHANGE 1 3
QUERY 1 2
DONE

Output:
1
3

分析：

对每个点建立一个结点，值为0

对每个边建立一个结点，值为边权。

然后用LCT维护即可

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<vector>
using namespace std;
#define maxn 500007
#define inf  1000000000
#define ll int

struct Node{
    Node *fa,*ch[2];
    bool rev,root;
    int val;
    ll minv;
};
Node pool[maxn];
Node *nil,*tree[maxn];
int cnt = 0;
void init(){
    cnt = 1;
    nil = tree[0] = pool;
    nil->ch[0] = nil->ch[1] = nil;
    nil->val = 0;
    nil->minv = 0;
}
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].minv = val;
    return &pool[cnt++];
}

//左右子树反转******真正把结点变为根
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){
    if(x == nil) return ;
    x->minv = x->val;
    Node*y = x->ch[0];
    if(y->minv > x->minv)
        x->minv = y->minv;
    y = x->ch[1];
    if(y->minv > x->minv)
        x->minv = y->minv;
}

//splay下推信息******
void pushdown(Node *x){
    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,*z;
    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;
}
Node * find(Node *root){
    if(root->ch[0] == nil) return root;
    return find(root->ch[0]);
}
Node*road[maxn];
int main(){
    int n,q,t;
    Node*x,*y,*z;
    scanf("%d",&t);
    char word[20];
    while(t--){
        scanf("%d",&n);

        init();
        int u,v,t;
        for(int i = 1;i <= n; i++)
            tree[i] = newnode(0,nil);
        for(int i = 1;i < n ;i++){
            scanf("%d%d%d",&u,&v,&t);
            road[i] = x = newnode(t,nil);
            link(tree[u],x);
            link(tree[v],x);
        }
        while(1){
            scanf("%s",word);
            if(word[0] == 'D') break;
            scanf("%d%d",&u,&v);
            if(word[0] == 'Q'){
                be_root(tree[u]);
                y = access(tree[v]);
                printf("%d\n",y->minv);
            }
            else {
                splay(road[u]);
                road[u]->val = v;
                update(road[u]);
            }
        }
    }

    return 0;
}

/*
3 100 1
1 2 1
2 3 2
0 2
1 2 3
0 3
1 2 5
0 2

*/