poj 3237 Tree(树链剖分)

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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;}
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