树链剖分

入门看这篇http://blog.sina.com.cn/s/blog_7a1746820100wp67.html

一个入门题目

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

代码：

#include <cstdio>
#include <cstring>
#include <iostream>
#define lson l, m, rt<<1
#define rson m+1, r, rt<<1|1 
using namespace std;

const int maxn = 10010;
struct Node{
    int to, next;
};
int p[maxn];//先线段树中的位置，dfs序 
int tree[maxn*4]; //线段树 
int fa[maxn];//保存当前节点的父亲节点 
int top[maxn];//表示当前节点所在链顶端的节点 
int siz[maxn];//当前节点的所有子节点数目 
int deep[maxn];//当前节点的深度 
int head[maxn];//前向星表示 
int son[maxn];//当前节点的重儿子 
Node edge[maxn*2];//前向星的表示法 
int e[maxn][3];//边集 
int tot, pos;//tot表示前向星中的所有节点个数，pos表示在线段树中的位置 

int Max(int a, int b)
{
    return a > b ? a : b;
}

void init()
{
    tot = 0;//初始化为0 
    pos = 1;//因为我写的线段树1是根节点，所以这个从1开始 
    memset(head, -1, sizeof(head));
    memset(son, -1, sizeof(son));    
}
//添加边 
void addedge(int u, int v)
{
    edge[tot].to = v; 
    edge[tot].next = head[u];
    head[u] = tot++;
}
//第一个dfs，求出来当前节点有多少个孩子，它的父亲，深度，还有重儿子 
void dfs1(int u, int pre, int d)
{
    siz[u] = 1;
    fa[u] = pre;
    deep[u] = d;
    
    for (int i = head[u]; i != -1; i = edge[i].next)//遍历所有的边 
    {
        int v = edge[i].to;
        if (v != pre)//不能向上走，因为这是无向图 
        {
            dfs1(v, u, d + 1);
            siz[u] += siz[v];
            if (son[u] == -1 || siz[son[u]] < siz[v])
                son[u] = v;        
        }
    }
}
//第二个dfs求出来top，和p 
void dfs2(int u, int sp)
{
    top[u] = sp;
    if (son[u] != -1)//不是叶子顶点 
    {
        p[u] = pos++;
        dfs2(son[u], sp);
    }
    else//叶子顶点 
    {
        p[u] = pos++;
        return;
    }
    for (int i = head[u]; i != -1; i = edge[i].next)//找它的连接的所有顶点 
    {
        int v = edge[i].to;
        if (son[u] != v && v != fa[u])//不是重儿子并且不能向上找 
        {
            dfs2(v, v);
        }
    } 
}
//线段树 
void pushup(int rt)
{
    tree[rt] = Max(tree[rt<<1], tree[rt<<1|1]);
}

void build(int l, int r, int rt)
{
    if (l == r)
    {
        tree[rt] = 0;
        return;
    }
    int m = (l + r) >> 1;
    build(lson);
    build(rson);
    pushup(rt); 
}
void update(int p, int sc, int l, int r, int rt)
{
    if (l == r)
    {
        tree[rt] = sc;
        return;
    }
    int m = (l + r) >> 1;
    if (p <= m)
        update(p, sc, lson);
    else
        update(p, sc, rson);
    pushup(rt); 
}

int query(int L, int R, int l, int r, int rt)
{
    if (L <= l && r <= R)
    {
        return tree[rt];
    }
    int m = (l + r) >> 1;
    int res = 0;
    if (L <= m)
        res = Max(res, query(L, R, lson));
    if (R > m)
        res = Max(res, query(L, R, rson));
    return res;
}
//查询u->v边的最大值 
int find(int u, int v)
{
    int f1 = top[u], f2 = top[v];
    int ans = 0;
    while (f1 != f2)
    {
        if (deep[f1] < deep[f2])
        {
            swap(f1, f2);
            swap(u, v);
        }
        ans = Max(ans, query(p[f1], p[u], 1, pos - 1, 1));
        u = fa[f1];
        f1 = top[u];
    }
    if (u == v)
        return ans;
    if (deep[u] > deep[v])
        swap(u, v);
    return Max(ans, query(p[son[u]], p[v], 1, pos - 1, 1));    
}

int main()
{
    //freopen("in.txt", "r", stdin);
    int T, n;
    scanf("%d", &T);
    while (T--)
    {
        init();
        scanf("%d", &n);
        for (int i = 0; i < n - 1; i++)
        {
            scanf("%d %d %d", &e[i][0], &e[i][1], &e[i][2]);
            addedge(e[i][0], e[i][1]);//无向图，所以添加两次 
            addedge(e[i][1], e[i][0]);
        }
        dfs1(1, 0, 0);
        dfs2(1, 1);
        build(1, pos - 1, 1);
        for (int i = 0; i < n - 1; i++)
        {
            if (deep[e[i][0]] > deep[e[i][1]])
                swap(e[i][0], e[i][1]);
            update(p[e[i][1]], e[i][2], 1, pos - 1, 1);//将权值更新到线段树中 
        }
        
        char op[10];
        int u, v;
        while (~scanf("%s", op))
        {
            if (op[0] == 'D')
                break;
            else if (op[0] == 'C')
            {
                scanf("%d %d", &u, &v);
                update(p[e[u - 1][1]], v, 1, pos - 1, 1);
            }
            else
            {
                scanf("%d %d", &u, &v);
                int t = find(u, v);
                printf("%d
", t);
            }
        }
    }    
    
    return 0;
}