AtCoder Beginner Contest 237 G - Range Sort Query
原题链接 G - Range Sort Query
思路:
\(x\)是固定的,所以考虑将排序维护成0 1串,为\(p_i < x\)为\(0\),那么\(p_i \geq x\)为\(1\),那么这样就可以用线段树维护了,我们可以维护每个区间1的个数,设为\(cnt\),对于升序来说,设我们需要更新的这个区间\([l, r]\),那么现在查询后,更新新的区间就是\([r - cnt + 1, 1]\)为\(1\),更新\([l, r - cnt]\)为\(0\),所以维护两颗线段树,一颗是\(p_i \geq x\),另一颗是\(p_i \geq x + 1\),那么最后可以从\(1 - N\)遍历查询,若\(t1.query(1, i, i)\)与\(t2.query(1, i, i)\)两者不相等,说明这就是我们要找的那个位置。
一开始把懒标记初值设成了0,怎么也不对,后来对拍了半天才发现有是因为懒标记有0 1值,那初值肯定不能是0 1,所以改成-1,就对了。
#include <bits/stdc++.h>
using namespace std;
const int N = 2E5 + 10;
int w[N], n, q, x;
struct SegMentTree {
struct node {
int l, r;
int val, tag;
}tr[N * 4];
void pushup(int u) {
tr[u].val = tr[u << 1].val + tr[u << 1 | 1].val;
}
void pushdown(int u) {
auto &root = tr[u], &left = tr[u << 1], &right = tr[u << 1 | 1];
if (~root.tag) {
left.tag = root.tag;
left.val = root.tag * (left.r - left.l + 1);
right.tag = root.tag;
right.val = root.tag * (right.r - right.l + 1);
root.tag = -1;
}
}
void build(int u, int l, int r, int x) {
tr[u] = {l, r};
tr[u].tag = -1;
if (l == r) {
tr[u].val = (w[l] >= x);
//tr[u].tag = 0;
return;
}
int mid = l + r >> 1;
build(u << 1, l, mid, x), build(u << 1 | 1, mid + 1, r, x);
pushup(u);
}
void modify(int u, int l, int r, int x) {
if (l <= tr[u].l && tr[u].r <= r) {
tr[u].tag = x;
tr[u].val = x * (tr[u].r - tr[u].l + 1);
return;
}
pushdown(u);
int mid = tr[u].l + tr[u].r >> 1;
if (l <= mid) modify(u << 1, l, r, x);
if (r > mid) modify(u << 1 | 1, l, r, x);
pushup(u);
}
int query(int u, int l, int r) {
if (l <= tr[u].l && tr[u].r <= r) {
return tr[u].val;
}
pushdown(u);
int mid = tr[u].l + tr[u].r >> 1;
int res = 0;
if (l <= mid) res += query(u << 1, l, r);
if (r > mid) res += query(u << 1 | 1, l, r);
return res;
}
}t1, t2;
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
//freopen("D:\\cpeditor\\competition\\Atcoder\\abc237\\data.txt", "r", stdin);
//freopen("D:\\cpeditor\\competition\\Atcoder\\abc237\\myans.txt", "w", stdout);
cin >> n >> q >> x;
for (int i = 1; i <= n; i++) cin >> w[i];
t1.build(1, 1, n, x);
t2.build(1, 1, n, x + 1);
while (q--) {
int op, l, r;
cin >> op >> l >> r;
int cnt1 = t1.query(1, l, r), cnt2 = t2.query(1, l, r);
int len = r - l + 1;
if (op == 1) { //asc
if (cnt1 && cnt1 != len) {
t1.modify(1, l, r - cnt1, 0); t1.modify(1, r - cnt1 + 1, r, 1);
}
if (cnt2 && cnt2 != len) {
t2.modify(1, l, r - cnt2, 0); t2.modify(1, r - cnt2 + 1, r, 1);
}
} else {
if (cnt1 && cnt1 != len) {
t1.modify(1, l, l + cnt1 - 1, 1); t1.modify(1, l + cnt1, r, 0);
}
if (cnt2 && cnt2 != len) {
t2.modify(1, l, l + cnt2 - 1, 1); t2.modify(1, l + cnt2, r, 0);
}
}
}
// cout << "test" << endl;
for(int i = 1; i <= n; i++) {
//cout << t1.query(1, i, i) << " " << t2.query(1, i, i) << endl;
if(t1.query(1, i, i) != t2.query(1, i, i)) {
cout << i << "\n";
}
}
//fclose(stdin);
//fclose(stdout);
return 0;
}