[洛谷P2590][ZJOI2008]树的统计

题目大意：一棵树，支持三个操作，

$CHANGE\;u\;t:$ 把结点$u$的权值改为$t$

$QMAX\;u\;v:$ 询问从点$u$到点$v$的路径上的节点的最大权值

$QSUM\;u\;v:$ 询问从点$u$到点$v$的路径上的节点的权值和

题解：裸的树链剖分

卡点：线段树区间修改我不知道那根筋搭错了，写了$l\;==\;r$（应为$L\;\leq\;l\;\&\&\;R\;\geq\;r$）

C++ Code：

#include <cstdio>
#define maxn 30010
using namespace std;
const int inf = 0x3f3f3f3f;
int n, m;
int w[maxn], V[maxn << 2], M[maxn << 2];
inline int max(int a, int b) {return a > b ? a : b;}
void swap(int &a, int &b) {a ^= b ^= a ^= b;}
void update(int rt) {
    V[rt] = V[rt << 1] + V[rt << 1 | 1];
    M[rt] = max(M[rt << 1], M[rt << 1 | 1]);
}
void build(int rt, int l, int r) {
    if (l == r) {
        V[rt] = M[rt] = w[l];
        return ;
    }
    int mid = l + r >> 1;
    build(rt << 1, l, mid);
    build(rt << 1 | 1, mid + 1, r);
    update(rt);
}
void add(int rt, int l, int r, int p, int num) {
    if (l == r) {
        V[rt] = M[rt] = num;
        return ;
    }
    int mid = l + r >> 1;
    if (p <= mid) add(rt << 1, l, mid, p, num);
    else add(rt << 1 | 1, mid + 1, r, p, num);
    update(rt);
}
int askS(int rt, int l, int r, int L, int R) {
    if (L <= l && R >= r) {
        return V[rt];
    }
    int mid = l + r >> 1, ans = 0;
    if (L <= mid) ans = askS(rt << 1, l, mid, L, R);
    if (R > mid) ans = ans + askS(rt << 1 | 1, mid + 1, r, L, R);
    return ans;
}
int askM(int rt, int l, int r, int L, int R) {
    if (L <= l && R >= r) {
        return M[rt];
    }
    int mid = l + r >> 1, ans = -inf;
    if (L <= mid) ans = askM(rt << 1, l, mid, L, R);
    if (R > mid) ans = max(ans, askM(rt << 1 | 1, mid + 1, r, L, R));
    return ans;
}

int head[maxn], cnt;
struct Edge {
    int to, nxt;
} e[maxn << 1];
void addE(int a, int b) {
    e[++cnt] = (Edge) {b, head[a]}; head[a] = cnt;
}
int fa[maxn], sz[maxn], son[maxn], dep[maxn];
int top[maxn], dfn[maxn], idx;
void dfs1(int rt) {
    sz[rt] = 1;
    for (int i = head[rt]; i; i = e[i].nxt) {
        int v = e[i].to;
        if (v != fa[rt]) {
            dep[v] = dep[rt] + 1;
            fa[v] = rt;
            dfs1(v);
            if (!son[rt] || sz[v] > sz[son[rt]]) son[rt] = v;
            sz[rt] += sz[v];
        }
    }
}
void dfs2(int rt) {
    dfn[rt] = ++idx;
    int v = son[rt];
    if (v) top[v] = top[rt], dfs2(v);
    for (int i = head[rt]; i; i = e[i].nxt) {
        v = e[i].to;
        if (v != son[rt] && v != fa[rt]) {
            top[v] = v;
            dfs2(v);
        }
    }
}
int queryS(int x, int y) {
    int ans = 0;
    while (top[x] != top[y]) {
        if (dep[top[x]] < dep[top[y]]) swap(x, y);
        ans += askS(1, 1, n, dfn[top[x]], dfn[x]);
        x = fa[top[x]];
    }
    if (dep[x] > dep[y]) swap(x, y);
    ans += askS(1, 1, n, dfn[x], dfn[y]);
    return ans;
}
int queryM(int x, int y) {
    int ans = -inf;
    while (top[x] != top[y]) {
        if (dep[top[x]] < dep[top[y]]) swap(x, y);
        ans = max(ans, askM(1, 1, n, dfn[top[x]], dfn[x]));
        x = fa[top[x]];
    }
    if (dep[x] > dep[y]) swap(x, y);
    ans = max(ans, askM(1, 1, n, dfn[x], dfn[y]));
    return ans;
}
int main() {
    scanf("%d", &n);
    for (int i = 1; i < n; i++) {
        int a, b;
        scanf("%d%d", &a, &b);
        addE(a, b);
        addE(b, a);
    }
    dep[1] = 1;
    dfs1(1);
    top[1] = 1;
    dfs2(1);
    for (int i = 1; i <= n; i++) {
        scanf("%d", &w[dfn[i]]);
    }
    build(1, 1, n);
    scanf("%d", &m);
    while (m --> 0) {
        char op[10];
        int x, y;
        scanf("%s%d%d", op, &x, &y);
        if (op[1] == 'M') {
            printf("%d\n", queryM(x, y));
        }
        if (op[1] == 'S') {
            printf("%d\n", queryS(x, y));
        }
        if (op[1] == 'H') {
            add(1, 1, n, dfn[x], y);
        }
    }
    return 0;
}