题目链接：poj 3237 Tree

题目大意：给定一棵树，三种操作：

• CHANGE i v：将i节点权值变为v
• NEGATE a b：将ab路径上所有节点的权值变为相反数
• QUERY a b：查询ab路径上节点权值的最大值。

解题思路：树链剖分，然后用线段树维护节点权值，成端更新查询。

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int maxn = 10005;
const int INF = 0x3f3f3f3f;

int id, far[maxn], son[maxn], cnt[maxn], idx[maxn], top[maxn], dep[maxn];

int N, en, first[maxn], jump[maxn * 2], val[maxn];
struct Edge {
    int u, v, w;
    void set (int u, int v, int w) {
        this->u = u;
        this->v = v;
        this->w = w;
    }
}ed[maxn * 2];

void dfs_fir (int u, int pre, int d) {
    dep[u] = d;
    far[u] = pre;
    cnt[u] = 1;
    son[u] = 0;

    for (int i = first[u]; i + 1; i = jump[i]) {
        int v = ed[i].v;
        if (v == pre)
            continue;
        dfs_fir(v, u, d + 1);
        cnt[u] += cnt[v];
        if (cnt[son[u]] < cnt[v])
            son[u] = v;
    }
}

void dfs_sec(int u, int rot) {
    idx[u] = id++;
    top[u] = rot;

    if (son[u])
        dfs_sec(son[u], rot);

    for (int i = first[u]; i + 1; i = jump[i]) {
        int v = ed[i].v;
        if (v == far[u] || v == son[u])
            continue;
        dfs_sec(v, v);
    }
}

#define lson(x) ((x)<<1)
#define rson(x) (((x)<<1)|1)

int lc[maxn << 2], rc[maxn << 2], filp[maxn << 2], Ma[maxn << 2], Mi[maxn << 2];

inline void maintain (int u) {
    filp[u] ^= 1;
    swap(Ma[u], Mi[u]);
    Ma[u] = -Ma[u];
    Mi[u] = -Mi[u];
}

inline void pushup(int u) {
    Ma[u] = max(Ma[lson(u)], Ma[rson(u)]);
    Mi[u] = min(Mi[lson(u)], Mi[rson(u)]);
}

inline void pushdown (int u) {
    if (filp[u]) {
        maintain(lson(u));
        maintain(rson(u));
        filp[u] = 0;
    }
}

void build (int u, int l, int r) {
    lc[u] = l;
    rc[u] = r;
    filp[u] = 0;

    if (l == r) {
        Ma[u] = Mi[u] = val[l];
        return;
    }

    int mid = (l + r) / 2;
    build(lson(u), l, mid);
    build(rson(u), mid + 1, r);
    pushup(u);
}

void modify (int u, int x, int w) {
    if (lc[u] == x && rc[u] == x) {
        Ma[u] = Mi[u] = w;
        return;
    }

    pushdown(u);
    int mid = (lc[u] + rc[u]) / 2;
    if (x <= mid)
        modify(lson(u), x, w);
    else
        modify(rson(u), x, w);
    pushup(u);
}

void splay(int u, int l, int r) {
    if (l <= lc[u] && rc[u] <= r) {
        maintain(u);
        return;
    }

    pushdown(u);
    int mid = (lc[u] + rc[u]) / 2;
    if (l <= mid)
        splay(lson(u), l, r);
    if (r > mid)
        splay(rson(u), l, r);
    pushup(u);
}

int query (int u, int l, int r) {
    if (l <= lc[u] && rc[u] <= r) 
        return Ma[u];

    pushdown(u);
    int mid = (lc[u] + rc[u]) / 2, ret = -INF;
    if (l <= mid)
        ret = max(ret, query(lson(u), l, r));
    if (r > mid)
        ret = max(ret, query(rson(u), l, r));
    pushup(u);
    return ret;
}

inline void add_Edge(int u, int v, int w) {
    ed[en].set(u, v, w);
    jump[en] = first[u];
    first[u] = en++;
}

void init () {
    en = 0;
    id = 1;
    memset(first, -1, sizeof(first));

    scanf("%d", &N);

    int u, v, w;
    for (int i = 0; i  < N - 1; i++) {
        scanf("%d%d%d", &u, &v, &w);
        add_Edge(u, v, w);
        add_Edge(v, u, w);
    }

    dfs_fir(1, 0, 0);
    dfs_sec(1, 1);


    for (int i = 0; i < N - 1; i++) {
        int t = i * 2;
        if (dep[ed[t].u] < dep[ed[t].v])
            swap(ed[t].u, ed[t].v);
        val[idx[ed[t].u]] = ed[t].w;
    }
    build(1, 1, N);
}

int query (int u, int v) {
    int p = top[u], q = top[v], ret = -INF;
    while (p != q) {
        if (dep[p] < dep[q]) {
            swap(p, q);
            swap(u, v);
        }
        ret = max(ret, query(1, idx[p], idx[u]));
        u = far[p];
        p = top[u];
    }
    //printf("%d %d\n", u, v);

    if (u == v)
        return ret;

    if (dep[u] > dep[v])
        swap(u, v);
    ret = max(ret, query(1, idx[son[u]], idx[v]));
    return ret;
}

void modify (int u, int v) {
    int p = top[u], q = top[v];
    while (p != q) {
        if (dep[p] < dep[q]) {
            swap(p, q);
            swap(u, v);
        }
        splay(1, idx[p], idx[u]);
        u = far[p];
        p = top[u];
    }

    if (u == v)
        return;

    if (dep[u] > dep[v])
        swap(u, v);
    splay(1, idx[son[u]], idx[v]);
}

void solve () {
    int u, v;
    char op[20];
    while (scanf("%s", op), strcmp(op, "DONE") != 0) {
        scanf("%d%d", &u, &v);
        if (op[0] == ‘C‘)
            modify(1, idx[ed[u*2-2].u], v);
        else if (op[0] == ‘Q‘)
            printf("%d\n", query(u, v));
        else
            modify(u, v);
    }
}

int main () {
    int cas;
    scanf("%d", &cas);
    while (cas--) {
        init();
        solve();
    }
    return 0;
}

poj 3237 Tree(树链剖分)