spoj 375. Query on a tree(树链剖分)

题目链接：spoj 375. Query on a tree

题目大意：

poj 3237的简化版，用同一份代码都能过。

解题思路：略。

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

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

#define lson(x) ((x)<<1)
#define rson(x) (((x)<<1)|1)
int val[maxn];
int lc[maxn << 2], rc[maxn << 2], s[maxn << 2];
inline void pushup(int u) {
    s[u] = max(s[lson(u)], s[rson(u)]);
}

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

    if (l == r) {
        s[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) {
        s[u] = w;
        return;
    }

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

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

int N, ne, first[maxn], jump[maxn * 2];
int id, idx[maxn], top[maxn], far[maxn], son[maxn], cnt[maxn], dep[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];

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

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

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

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

    if (son[u])
        dfs(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(v, v);
    }
}

void init () {
    int u, v, w;
    scanf("%d", &N);
    ne = id = 0;
    memset(first, -1, sizeof(first));
    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(1, 0, 0);
    dfs(1, 1);

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

int solve (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];
    }
    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;
}

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

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