LightOJ 1348 Aladdin and the Return Journey 树链剖分

题目：http://lightoj.com/login_main.php?url=volume_showproblem.php?problem=1348

题意：给定一个数，节点数为n。先输入一个n，代表节点数，然后一行有n个数，代表 树上点从0 到n - 1的初始点权，然后n - 1行，每行两个数a b，代表这两个点之间有一条边。紧接着输入一个m，代表有m个操作，操作有两种：0 a b询问点a到点b路径上的点的权值和，1 a b把点a的权值更新为b

思路：基本树链剖分模板题啦，把树剖分到线段树上解决，很简单，另外此题中点从0开始，所以所有出现点的地方我都加了1，因为我用的板子是从1开始的

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int N = 30010;
struct edge
{
    int to, next;
} g[N*2];
struct node
{
    int l, r, val;
} s[N*4];
int head[N], siz[N], son[N], top[N], id[N], dep[N], fat[N];
int val[N], tmp[N];
int n, m, cnt, num, _case = 0;
void add_edge(int v, int u)
{
    g[cnt].to = u;
    g[cnt].next = head[v];
    head[v] = cnt++;
}
void dfs1(int v, int d, int fa)
{
    dep[v] = d, siz[v] = 1, son[v] = 0, fat[v] = fa;
    for(int i = head[v]; i != -1; i = g[i].next)
    {
        int u = g[i].to;
        if(u != fa)
        {
            dfs1(u, d + 1, v);
            siz[v] += siz[u];
            if(siz[son[v]] < siz[u]) son[v] = u;
        }
    }
}
void dfs2(int v, int tp)
{
    top[v] = tp, id[v] = ++num;
    if(son[v]) dfs2(son[v], tp);
    for(int i = head[v]; i != -1; i = g[i].next)
    {
        int u = g[i].to;
        if(u != fat[v] && u != son[v]) dfs2(u, u);
    }
}
void push_up(int k)
{
    s[k].val = s[k<<1].val + s[k<<1|1].val;
}
void build(int l, int r, int k)
{
    s[k].l = l, s[k].r = r;
    if(l == r)
    {
        s[k].val = val[l];
        return;
    }
    int mid = (l + r) >> 1;
    build(l, mid, k << 1);
    build(mid + 1, r, k << 1|1);
    push_up(k);
}
void update(int x, int c, int k)
{
    if(s[k].l == s[k].r)
    {
        s[k].val = c;
        return;
    }
    int mid = (s[k].l + s[k].r) >> 1;
    if(x <= mid) update(x, c, k << 1);
    else update(x, c, k << 1|1);
    push_up(k);
}
int query(int l, int r, int k)
{
    if(l <= s[k].l && s[k].r <= r)
        return s[k].val;
    int mid = (s[k].l + s[k].r) >> 1;
    int ans = 0;
    if(l <= mid) ans += query(l, r, k << 1);
    if(r > mid) ans += query(l, r, k << 1|1);
    return ans;
}
int seek(int v, int u)
{
    int t1 = top[v], t2 = top[u], ans = 0;
    while(t1 != t2)
    {
        if(dep[t1] < dep[t2])
            swap(t1, t2), swap(v, u);
        ans += query(id[t1], id[v], 1);
        v = fat[t1], t1 = top[v];
    }
    if(dep[v] > dep[u]) swap(v, u);
    return ans + query(id[v], id[u], 1);
}
void slove()
{
    int a, b, c;

    scanf("%d", &m);
    printf("Case %d:\n", ++_case);
    for(int i = 1; i <= m; i++)
    {
        scanf("%d%d%d", &a, &b, &c);
        if(a == 0)
        {
            b++, c++;
            printf("%d\n", seek(b, c));
        }
        else
        {
            b++;
            update(id[b], c, 1);
        }
    }
}
int main()
{
    int t, a, b;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d", &n);
        cnt = num = 0;
        memset(head, -1, sizeof head);
        for(int i = 1; i <= n; i++)
            scanf("%d", tmp + i);
        for(int i = 1; i <= n - 1; i++)
        {
            scanf("%d%d", &a, &b);
            a++, b++;
            add_edge(a, b);
            add_edge(b, a);
        }
        dfs1(1, 1, 0);
        dfs2(1, 1);
        for(int i = 1; i <= n; i++)
            val[id[i]] = tmp[i];
        build(1, num, 1);
        slove();
    }
    return 0;
}