HYSBZ 1036(树链剖分)

题意：有一棵树，有n个点，n - 1条边，给出每个节点的权值，有三种操作，qmax a b询问a到b所有点的最大权值，qsum a b询问a到b所有点权值和，change a b把节点a权值设为b。
题解：两颗线段树，维护最大值和区间和。

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N = 30005;const int INF = 0x3f3f3f3f;struct Edge {    int u, v, nxt;    Edge() {}    Edge(int a, int b, int c): u(a), v(b), nxt(c) {}}e[N << 1];int size[N], top[N], son[N], fa[N], dep[N], id[N];int n, q, cnt, tot, head[N], val[N], tree[N << 2], sum[N << 2];void AddEdge(int u, int v) {    e[cnt] = Edge(u, v, head[u]);    head[u] = cnt++;    e[cnt] = Edge(v, u, head[v]);    head[v] = cnt++;}void dfs2(int u, int tp) {    top[u] = tp;    id[u] = ++tot;    if (son[u]) dfs2(son[u], tp);    for (int i = head[u]; i + 1; i = e[i].nxt) {        int v = e[i].v;        if (v == fa[u] || v == son[u]) continue;        dfs2(v, v);    }}void dfs1(int u, int f, int depth) {    size[u] = 1, son[u] = 0, fa[u] = f, dep[u] = depth + 1;    for (int i = head[u]; i + 1; i = e[i].nxt) {        int v = e[i].v;         if (v == f) continue;        dfs1(v, u, depth + 1);        size[u] += size[v];        if (size[son[u]] < size[v]) son[u] = v;    }}void pushup(int k) {    tree[k] = max(tree[k * 2], tree[k * 2 + 1]);    sum[k] = sum[k * 2] + sum[k * 2 + 1];}void modify(int k, int left, int right, int pos, int v) {    if (left == right) {        tree[k] = sum[k] = v;        return;    }    int mid = (left + right) / 2;    if (pos <= mid)        modify(k * 2, left, mid, pos, v);    else        modify(k * 2 + 1, mid + 1, right, pos, v);    pushup(k);}int query(int k, int left, int right, int l, int r, int flag) {    if (l <= left && right <= r) {        if (!flag) return tree[k];        return sum[k];    }    int mid = (left + right) / 2, res;    if (!flag) res = -INF;    else res = 0;    if (l <= mid) {        if (!flag) res = max(res, query(k * 2, left, mid, l, r, flag));        else res += query(k * 2, left, mid, l, r, flag);    }    if (r > mid) {        if (!flag) res = max(res, query(k * 2 + 1, mid + 1, right, l, r, flag));        else res += query(k * 2 + 1, mid + 1, right, l, r, flag);    }    return res;}void solve(int u, int v, int flag) {    int tp1 = top[u], tp2 = top[v], res;    if (!flag) res = -INF;    else res = 0;    while (tp1 != tp2) {        if (dep[tp1] < dep[tp2]) {            swap(tp1, tp2);            swap(u, v);        }        if (!flag) res = max(res, query(1, 1, tot, id[tp1], id[u], 0));        else res += query(1, 1, tot, id[tp1], id[u], 1);        u = fa[tp1];        tp1 = top[u];    }    if (dep[u] > dep[v]) swap(u, v);    if (!flag) res = max(res, query(1, 1, tot, id[u], id[v], 0));    else res += query(1, 1, tot, id[u], id[v], 1);    printf("%d\n", res);}int main() {    while (scanf("%d", &n) == 1) {        int u, v;        cnt = tot = 0;        memset(head, -1, sizeof(head));        memset(tree, 0, sizeof(tree));        memset(sum, 0, sizeof(sum));        for (int i = 1; i < n; i++) {            scanf("%d%d", &u, &v);            AddEdge(u, v);        }        for (int i = 1; i <= n; i++)            scanf("%d", &val[i]);        dfs1(1, 0, 1);        dfs2(1, 1);        for (int i = 1; i <= n; i++)            modify(1, 1, tot, id[i], val[i]);        scanf("%d", &q);        char op[10];        int a, b;        while (q--) {            scanf("%s%d%d", op, &a, &b);            if (op[1] == 'M')                solve(a, b, 0);            else if (op[1] == 'S')                solve(a, b, 1);            else                modify(1, 1, tot, id[a], b);        }    }    return 0;}