SPOJ - QTREE Query on a tree(树剖)

最新推荐文章于 2019-06-02 11:55:24 发布

原创最新推荐文章于 2019-06-02 11:55:24 发布 · 554 阅读

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本文深入探讨了树剖裸题的概念及其解决方法，提供了详细的解题思路和代码实现，包括如何构建树状数组、进行区间更新和查询最大值。通过实例演示，帮助读者理解并掌握这一经典算法。

题目大意：给出一棵树，有两种操作
CHANGE i t：将第i条边的权值变成t
QUERY a b：询问a到b的路径上权值最大的边

解题思路：树剖裸题

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 10010;
const int M = 10010 << 2;

int n;
struct Edge{
    int u, v, dis, next;
    Edge() {}
    Edge(int u, int v, int dis, int next): u(u), v(v), dis(dis), next(next) {}
}E[2 * N];

int head[N], top[N], son[N], size[N], id[N], f[N], dep[N], val[N], Max[M];
int tot, idx;

void build(int u, int l, int r) {
    if (l == r) {
        Max[u] = val[l];
        return ;
    }
    int mid = (l + r) >> 1;
    build(2 * u, l, mid);
    build(2 * u + 1, mid + 1, r);
    Max[u] = max(Max[2 * u], Max[2 * u + 1]);
}

void dfs1(int u, int fa, int depth) {
    dep[u] = depth; f[u] = fa; size[u] = 1; son[u] = 0;

    for (int i = head[u]; ~i; i = E[i].next) {
        int v = E[i].v;
        if (v == fa) continue;
        dfs1(v, u, depth + 1);
        size[u] += size[v];
        if (size[son[u]] < size[v]) son[u] = v;
    }
}

void dfs2(int u, int tp) {
    top[u] = tp;
    id[u] = ++idx;
    if (son[u]) dfs2(son[u], tp);

    for (int i = head[u]; ~i; i = E[i].next) {
        int v = E[i].v;
        if (v == f[u] || v == son[u]) continue;
        dfs2(v, v);
    }
}

int query(int u, int l, int r, int L, int R) {
    if (L <= l && r <= R)  return Max[u];
    int mid = (l + r) >> 1;
    int ans = 0;
    if (L <= mid) ans = max(ans, query(2 * u, l, mid, L, R));
    if (R > mid) ans = max(ans, query(2 * u + 1, mid + 1, r, L, R));
    return ans;
}

void modify(int u, int l, int r, int pos, int c) {
    if (l == r && l == pos) {
        Max[u] = c;
        return ;
    }
    int mid = (l + r) >> 1;
    if (pos <= mid) modify(2 * u, l, mid, pos, c);
    else modify(2 * u + 1, mid + 1, r, pos, c);
    Max[u] = max(Max[u * 2], Max[2 * u + 1]);
}

int gao(int u, int v) {
    int tp1 = top[u], tp2 = top[v];
    int ans = 0;
    while (tp1 != tp2) {
        if (dep[tp1] < dep[tp2]) {
            swap(tp1, tp2);
            swap(u, v);
        }
        ans = max(ans, query(1, 1, idx, id[tp1], id[u]));
        u = f[tp1];
        tp1 = top[u];
    }
    if (u == v) return ans;
    if (dep[u] > dep[v]) swap(u, v);
    ans = max(ans, query(1, 1, idx, id[son[u]], id[v]));
    return ans;
}


void solve() {
    char op[30];
    int a, b;
    while (scanf("%s", op) && op[0] != 'D') {
        scanf("%d%d", &a, &b);

        if (op[0] == 'Q')  printf("%d\n", gao(a, b));
        else {
            a--;
            modify(1, 1, idx, id[E[a * 2].u], b);
        }
    }
}

void AddEdge(int u, int v, int dis) {
    E[tot] = Edge(u, v, dis, head[u]);
    head[u] = tot++;
    E[tot] = Edge(v, u, dis, head[v]);
    head[v] = tot++;
}

void init() {
    scanf("%d", &n);
    tot = 0;
    memset(head, -1, sizeof(head));

    int u, v, dis;
    for (int i = 1; i < n; i++)  {
        scanf("%d%d%d", &u, &v, &dis);
        AddEdge(u, v, dis);
    }

    idx = 0;
    dfs1(1, 0, 1);
    dfs2(1, 1);

    for (int i = 0; i < tot; i += 2) {
        if (dep[E[i].u] < dep[E[i].v]) swap(E[i].u, E[i].v);
        val[id[E[i].u]] = E[i].dis;
    }
    build(1, 1, idx);
}

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