BZOJ4034【树链剖分】【线段树】
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/* I will wait for you */ #include <cstdio> #include <cstdlib> #include <cstring> #include <cmath> #include <ctime> #include <algorithm> #include <iostream> #include <fstream> #include <vector> #include <queue> #include <deque> #include <set> #include <map> #include <string> #define make(a,b) make_pair(a,b) #define fi first #define se second using namespace std; typedef long long ll; typedef unsigned long long ull; typedef pair<int, int> pii; typedef map<int, int> mii; const int maxn = 100010; const int maxm = 1010; const int maxs = 26; const int inf = 0x3f3f3f3f; const int P = 1000000007; const double error = 1e-9; inline ll read() { ll x = 0, f = 1; char ch = getchar(); while (ch > '9' || ch < '0' ) f = (ch == '-' ? -1 : 1), ch = getchar(); while (ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar(); return f * x; } struct edge { int v, next; } e[2 * maxn]; struct node { int l, r; ll sum, add; } t[4 * maxn]; int n, m, cnt, dfn, per[maxn], fa[maxn], mx[maxn], pr[maxn],deep[maxn], size[maxn], pos[maxn], head[maxn]; void insert(int u, int v) { e[cnt] = (edge) {v, head[u]}, head[u] = cnt++; e[cnt] = (edge) {u, head[v]}, head[v] = cnt++; } void dfs1(int u) { size[u] = 1; for (int i = head[u]; i != -1; i = e[i].next) { int v = e[i].v; if (v != fa[u]) { fa[v] = u, deep[v] = deep[u] + 1; dfs1(v), size[u] += size[v]; if (size[v] > size[mx[u]]) mx[u] = v; } } } void dfs2(int u, int p) { pos[u] = ++dfn, per[u] = p; if (mx[u]) dfs2(mx[u], p); for (int i = head[u]; i != -1; i = e[i].next) { int v = e[i].v; if (v != fa[u] && v != mx[u]) dfs2(v, v); } } void build(int u, int l, int r) { t[u].l = l, t[u].r = r; if (l != r) { int mid = (l + r) / 2; build(2 * u, l, mid); build(2 * u + 1, mid + 1, r); } } void maintain(int u){int l = t[u].l, r = t[u].r;t[u].sum = 0;if (r > l)t[u].sum = t[2 * u].sum + t[2 * u + 1].sum;t[u].sum += (r - l + 1) * t[u].add;}void add(int u, int x, int y, int c){int l = t[u].l, r = t[u].r;if (l >= x && r <= y)t[u].add += c;else {int mid = (l + r) / 2;if (x <= mid)add(2 * u, x, y, c);if (y > mid)add(2 * u + 1, x, y, c);}maintain(u);}ll query(int u, int x, int y, ll add){int l = t[u].l, r = t[u].r;if (l >= x && r <= y)return t[u].sum + (r - l + 1) * add;else {int mid = (l + r) / 2; ll ans = 0;if (x <= mid)ans += query(2 * u, x, y, add + t[u].add);if (y > mid)ans += query(2 * u + 1, x, y, add + t[u].add);return ans;}}ll query(int u){ll ans = 0;while (u) {ans += query(1, pos[per[u]], pos[u], 0);u = fa[per[u]];}return ans;}int main() { n = read(), m = read(), build(1, 1, n); memset(head, -1, sizeof head); for (int i = 1; i <= n; i++) pr[i] = read(); for (int i = 1; i < n; i++) { int u = read(), v = read(); insert(u, v); } dfs1(1), dfs2(1, 1); for (int i = 1; i <= n; i++) add(1, pos[i], pos[i], pr[i]); for (int i = 1; i <= m; i++) { int t = read(); if (t == 1) { int u = read(), c = read(); add(1, pos[u], pos[u], c); } if (t == 2) { int u = read(), c = read(); add(1, pos[u], pos[u] + size[u] - 1, c); } if (t == 3) { int u = read(); printf("%lld\n", query(u)); } } return 0; }
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- BZOJ4034【树链剖分】【线段树】
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