LightOJ - 1348 Aladdin and the Return Journey(树剖)

题目大意：给出一棵树和两种操作
0 i j：从i点到j点的权值和
1 i v: 第i点的权值变成v

解题思路：树剖裸题

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N = 30010;const int M = 30010 << 2;struct Edge{    int u, v, next;    Edge() {}    Edge(int u, int v, int next): u(u), v(v), next(next) {}}E[N * 2];int size[N], son[N], top[N], id[N], head[N], val[N], id2[N], dep[N], f[N];int left[M], right[M], sum[M];int idx, n, tot, cas = 1;void build(int u, int l, int r) {    left[u] = l; right[u] = r;    if (l == r) {        sum[u] = val[id2[l]];        return ;    }    int mid = (l + r) >> 1;    build(2 * u, l, mid);    build(2 * u + 1, mid + 1, r);    sum[u] = sum[2 * u] + sum[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);    }}void AddEdge(int u, int v) {    E[tot] = Edge(u, v, head[u]);    head[u] = tot++;    E[tot] = Edge(v, u, head[v]);    head[v] = tot++;}int query(int u, int l, int r) {    if (l <= left[u] && right[u] <= r) return sum[u];    int mid = (left[u] + right[u]) >> 1;    int ans = 0;    if (l <= mid) ans += query(2 * u, l, r);    if (r > mid) ans +=  query(2 * u + 1, l, r);    return ans;}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 += query(1, id[tp1], id[u]);        u = f[tp1];        tp1 = top[u];    }    if (dep[u] > dep[v]) swap(u, v);    ans += query(1, id[u], id[v]);    return ans;}void modify(int u, int pos, int c) {    if (left[u] == right[u] && left[u] == pos) {        sum[u] = c;        return ;    }    int mid = (left[u] + right[u]) >> 1;    if (pos <= mid) modify(2 * u, pos, c);    else modify(2 * u + 1, pos, c);     sum[u] = sum[2 * u + 1] + sum[2 * u];}void solve() {    printf("Case %d:\n", cas++);    int q, op, a, b;    scanf("%d", &q);    while (q--) {        scanf("%d%d%d", &op, &a, &b);        if (op) modify(1, id[a + 1], b);        else printf("%d\n", gao(a + 1, b + 1));    }}void init() {    scanf("%d", &n);    memset(head, -1, sizeof(head));    tot = 0;    for (int i = 1; i <= n; i++)        scanf("%d", &val[i]);    int u, v;    for (int i = 1; i < n; i++) {        scanf("%d%d", &u, &v);        AddEdge(u + 1, v + 1);    }    idx = 0;    dfs1(1, 0, 1);    dfs2(1, 1);    for (int i = 1; i <= n; i++) id2[id[i]] = i;    build(1, 1, idx);}int main() {    int test;    scanf("%d", &test);    while (test--) {        init();        solve();    }    return 0;}