树链剖分 【知识点】+【模板】

知识点讲解链接

简单模版
QTREE - Query on a tree
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

son[]记录重儿子，dep[]记录深度，num[]记录孩子个数，pre[]记录父节点。
top[]记录重链头节点，pos[]记录节点与父亲连边在线段树中的位置。
更新时直接更新，保证u -> v的连边u是父亲。
查询时

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define INF 0x3f3f3f3f
#define eps 1e-8
#define MAXN (10000+10)
#define MAXM (300000+10)
#define Ri(a) scanf("%d", &a)
#define Rl(a) scanf("%lld", &a)
#define Rf(a) scanf("%lf", &a)
#define Rs(a) scanf("%s", a)
#define Pi(a) printf("%d\n", (a))
#define Pf(a) printf("%.2lf\n", (a))
#define Pl(a) printf("%lld\n", (a))
#define Ps(a) printf("%s\n", (a))
#define W(a) while((a)--)
#define CLR(a, b) memset(a, (b), sizeof(a))
#define MOD 1000000007
#define LL long long
#define lson o<<1, l, mid
#define rson o<<1|1, mid+1, r
#define ll o<<1
#define rr o<<1|1
#define PI acos(-1.0)
#pragma comment(linker, "/STACK:102400000,102400000")
#define fi first
#define se second
using namespace std;
struct Tree{
    int l, r, Max;
};
Tree tree[MAXN<<2];
void PushUp(int o){
    tree[o].Max = max(tree[ll].Max, tree[rr].Max);
}
void Build(int o, int l, int r)
{
    tree[o].l = l; tree[o].r = r;
    tree[o].Max = 0;
    if(l == r)
        return ;
    int mid = (l + r) >> 1;
    Build(lson); Build(rson);
}
void Update(int o, int pos, int v)
{
    if(tree[o].l == tree[o].r)
    {
        tree[o].Max = v;
        return ;
    }
    int mid = (tree[o].l + tree[o].r) >> 1;
    if(pos <= mid) Update(ll, pos, v);
    else Update(rr, pos, v);
    PushUp(o);
}
int Query(int o, int L, int R)
{
    if(tree[o].l == L && tree[o].r == R)
        return tree[o].Max;
    int mid = (tree[o].l + tree[o].r) >> 1;
    if(R <= mid) return Query(ll, L, R);
    else if(L > mid) return Query(rr, L, R);
    else return max(Query(ll, L, mid), Query(rr, mid+1, R));
}
struct Edge{
    int from, to, val, next;
};
Edge edge[MAXN<<1];
int head[MAXN], edgenum;
int s[MAXN], e[MAXN], c[MAXN];
void init(){
    edgenum = 0; CLR(head, -1);
}
void addEdge(int u, int v, int w)
{
    Edge E = {u, v, w, head[u]};
    edge[edgenum] = E;
    head[u] = edgenum++;
}
int son[MAXN], num[MAXN];
int top[MAXN], pos[MAXN], id;
int dep[MAXN], pre[MAXN];
void DFS1(int u, int fa, int d)
{
    dep[u] = d; pre[u] = fa; num[u] = 1; son[u] = -1;
    for(int i = head[u]; i != -1; i = edge[i].next)
    {
        int v = edge[i].to;
        if(v == fa) continue;
        DFS1(v, u, d+1);
        num[u] += num[v];
        if(son[u] == -1 || num[son[u]] < num[v])
            son[u] = v;
    }
}
void DFS2(int u, int T)
{
    top[u] = T; pos[u] = ++id;
    if(son[u] == -1) return ;
    DFS2(son[u], T);
    for(int i = head[u]; i != -1; i = edge[i].next)
    {
        int v = edge[i].to;
        if(v == pre[u] || v == son[u]) continue;
        DFS2(v, v);
    }
}
int GetMax(int u, int v)
{
    int f1 = top[u], f2 = top[v];
    int ans = 0;
    while(f1 != f2)
    {
        if(dep[f1] < dep[f2])
        {
            swap(u, v);
            swap(f1, f2);
        }
        ans = max(ans, Query(1, pos[f1], pos[u]));
        u = pre[f1], f1 = top[u];
    }
    if(u == v) return ans;
    if(dep[u] > dep[v]) swap(u, v);
    return max(ans, Query(1, pos[son[u]], pos[v]));
}
int main()
{
    int t; Ri(t);
    W(t)
    {
        int n; Ri(n); init();
        for(int i = 1; i <= n-1; i++)
        {
            Ri(s[i]), Ri(e[i]), Ri(c[i]);
            addEdge(s[i], e[i], c[i]);
            addEdge(e[i], s[i], c[i]);
        }
        DFS1(1, -1, 1); id = 0; DFS2(1, 1); Build(1, 1, id);
        for(int i = 1; i <= n-1; i++)
        {
            if(dep[s[i]] > dep[e[i]])
                swap(s[i], e[i]);
            Update(1, pos[e[i]], c[i]);
        }
        char str[10];
        while(Rs(str), strcmp(str, "DONE"))
        {
            int x, y; Ri(x); Ri(y);
            if(str[0] == 'Q')
                Pi(GetMax(x, y));
            else
                Update(1, pos[e[x]], y);
        }
    }
    return 0;
}