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洛谷P2042 [NOI2005] 维护数列 题解 splay tree

题目大意:支持如下几个操作:

  1. 插入:​​INSERT p tot c1 c2 ... ctot​​:在第 \(p\) 个数后面插入连续的 \(tot\)
  2. 删除:​​DELETE p tot​​:删除第 \(p\) 个数开始的连续 \(tot\)
  3. 修改:​​MAKE-SAME p tot c​​:将第 \(p\) 个数开始的连续 \(tot\) 个数字全部更新为 \(c\)
  4. 翻转:​​REVERSE p tot​​:将第 \(p\) 个数开始的连续 \(tot\) 个数的区间(即:\([p, p+tot)\))进行翻转
  5. 求和:​​GET-SUM p tot​​:求第 \(p\) 个数开始的连续 \(tot\)
  6. 求最大连续子序列和:​​MAX-SUM​​:求(整个)序列的最大连续子序列和

示例程序:

#include <bits/stdc++.h>
using namespace std;
const int maxn = 5e5 + 5, INF = 1e9;

struct Node {
int s[2], p, v;
int rev, same;
int sz, sum, ms, ls, rs;
Node() {}
Node(int _v, int _p) {
s[0] = s[1] = 0, v= _v, p = _p;
rev = same = 0;
sz = 1, sum = ms = v;
ls = rs = max(v, 0);
}
} tr[maxn];
int root;
int w[maxn];
queue<int> nodes;

void push_up(int x) {
auto &u = tr[x], &l = tr[u.s[0]], &r = tr[u.s[1]];
u.sz = l.sz + r.sz + 1;
u.sum = l.sum + r.sum + u.v;
u.ls = max(l.ls, l.sum + u.v + r.ls);
u.rs = max(r.rs, l.rs + u.v + r.sum);
u.ms = max(max(l.ms, r.ms), l.rs + u.v + r.ls);
}

void t_same(int x, int c) {
if (x) {
tr[x].same = 1;
tr[x].rev = 0;
tr[x].v = c;
tr[x].sum = tr[x].sz * c;
if (c > 0)
tr[x].ms = tr[x].ls = tr[x].rs = tr[x].sum;
else
tr[x].ms = c, tr[x].ls = tr[x].rs = 0;
}
}

void t_rev(int x) {
if (x) {
tr[x].rev ^= 1;
swap(tr[x].s[0], tr[x].s[1]);
swap(tr[x].ls, tr[x].rs);
}
}

void push_down(int x) {
if (tr[x].same) {
tr[x].same = tr[x].rev = 0;
t_same(tr[x].s[0], tr[x].v);
t_same(tr[x].s[1], tr[x].v);
}
else if (tr[x].rev) {
tr[x].rev = 0;
t_rev(tr[x].s[0]);
t_rev(tr[x].s[1]);
}
}

void f_s(int p, int u, bool k) {
tr[p].s[k] = u;
tr[u].p = p;
}

void rot(int x) {
int y = tr[x].p, z = tr[y].p;
bool k = tr[y].s[1] == x;
f_s(z, x, tr[z].s[1]==y);
f_s(y, tr[x].s[k^1], k);
f_s(x, y, k^1);
push_up(y), push_up(x);
}

void splay(int x, int k) {
while (tr[x].p != k) {
int y = tr[x].p, z = tr[y].p;
if (z != k)
(tr[y].s[1]==x) ^ (tr[z].s[1]==y) ? rot(x) : rot(y);
rot(x);
}
if (!k) root = x;
}

int get_k(int k) {
int u = root;
while (u) {
push_down(u);
if (tr[tr[u].s[0]].sz >= k) u = tr[u].s[0];
else if (tr[tr[u].s[0]].sz + 1 == k) return u;
else k -= tr[tr[u].s[0]].sz + 1, u = tr[u].s[1];
}
return -1;
}

int build(int l, int r, int p) {
int mid = (l + r) / 2;
int u = nodes.front(); nodes.pop();
tr[u] = Node(w[mid], p);
if (l < mid) tr[u].s[0] = build(l, mid-1, u);
if (mid < r) tr[u].s[1] = build(mid+1, r, u);
push_up(u);
return u;
}

void del(int u) {
if (!u) return;
nodes.push(u);
del(tr[u].s[0]);
del(tr[u].s[1]);
}

void init() {
// 初始化 tr[0]
tr[0] = Node(0, 0);
tr[0].sz = 0;
tr[0].ms = -INF;
// 把所有点放进回收站
for (int i = 1; i < maxn; i++) nodes.push(i);
}

int n, m;
char op[22];

int main() {
init();
scanf("%d%d", &n, &m);
w[0] = w[n+1] = -INF; // 设立两个哨兵节点
for (int i = 1; i <= n; i++) scanf("%d", w+i);
root = build(0, n+1, 0);
while (m--) {
scanf("%s", op);
if (!strcmp(op, "INSERT")) {
int p, tot;
scanf("%d%d", &p, &tot);
for (int i = 0; i < tot; i++) scanf("%d", w+i);
int l = get_k(p+1), r = get_k(p+2);
splay(l, 0), splay(r, l);
int u = build(0, tot-1, r);
tr[r].s[0] = u;
push_up(r), push_up(l);
}
else if (!strcmp(op, "DELETE")) {
int p, tot;
scanf("%d%d", &p, &tot);
int l = get_k(p), r = get_k(p+tot+1);
splay(l, 0), splay(r, l);
del(tr[r].s[0]);
tr[r].s[0] = 0;
push_up(r), push_up(l);
}
else if (!strcmp(op, "MAKE-SAME")) {
int p, tot, c;
scanf("%d%d%d", &p, &tot, &c);
int l = get_k(p), r = get_k(p+tot+1);
splay(l, 0), splay(r, l);
t_same(tr[r].s[0], c);
push_up(r), push_up(l);
}
else if (!strcmp(op, "REVERSE")) {
int p, tot;
scanf("%d%d", &p, &tot);
int l = get_k(p), r = get_k(p+tot+1);
splay(l, 0), splay(r, l);
t_rev(tr[r].s[0]);
push_up(r), push_up(l);
}
else if (!strcmp(op, "GET-SUM")) {
int p, tot;
scanf("%d%d", &p, &tot);
int l = get_k(p), r = get_k(p+tot+1);
splay(l, 0), splay(r, l);
printf("%d\n", tr[tr[r].s[0]].sum);
}
else { // MAX-SUM
printf("%d\n", tr[root].ms);
}
}
return 0;
}



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