P6492 STEP

考点

  • 线段树

题解

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#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 50;
int n, q, a[maxn];

struct {
int l, r;
int lx, rx, mx;
#define l(x) tr[x].l
#define r(x) tr[x].r
#define lx(x) tr[x].lx
#define rx(x) tr[x].rx
#define mx(x) tr[x].mx
#define lc(x) x * 2
#define rc(x) x * 2 + 1
} tr[4 * maxn];

void up(int p, int mid) {
if (a[mid] != a[mid + 1]) {
mx(p) = max(max(mx(lc(p)), mx(rc(p))), rx(lc(p)) + lx(rc(p)));
lx(p) = mx(lc(p)) == r(lc(p)) - l(lc(p)) + 1 ? mx(lc(p)) + lx(rc(p))
: lx(lc(p));
rx(p) = mx(rc(p)) == r(rc(p)) - l(rc(p)) + 1 ? mx(rc(p)) + rx(lc(p))
: rx(rc(p));
} else {
mx(p) = max(mx(lc(p)), mx(rc(p)));
lx(p) = lx(lc(p)), rx(p) = rx(rc(p));
}
}

void build(int p, int l, int r) {
l(p) = l, r(p) = r;
if (l == r) {
lx(p) = rx(p) = mx(p) = 1;
return;
}
int mid = (l + r) / 2;
build(lc(p), l, mid), build(rc(p), mid + 1, r);
up(p, mid);
}

void update(int p, int x) {
if (l(p) == x && r(p) == x) {
a[x] ^= 1;
return;
}
int mid = (l(p) + r(p)) / 2;
if (x <= mid) update(lc(p), x);
if (x > mid) update(rc(p), x);
up(p, mid);
}

int main() {
scanf("%d%d", &n, &q);
int x;
build(1, 1, n);
while (q--) {
scanf("%d", &x);
update(1, x);
printf("%d\n", mx(1));
}
return 0;
}

思路

考虑线段树。

最长满足要求的子串,要么经过中间点mid,要么在左孩子,要么在右孩子。

lx(p)代表当前节点p的区间从左边界向右走,能满足要求的最长连续子串长度;

rx(p)代表当前节点p的区间从右边界向左走,能满足要求的最长连续子串长度。

如果a[mid] != a[mid + 1],说明左孩子与右孩子可以有交集,很容易得到下面代码:

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mx(p) = max(max(mx(lc(p)), mx(rc(p))), rx(lc(p)) + lx(rc(p)));
lx(p) = mx(lc(p)) == r(lc(p)) - l(lc(p)) + 1 ? mx(lc(p)) + lx(rc(p))
: lx(lc(p));
rx(p) = mx(rc(p)) == r(rc(p)) - l(rc(p)) + 1 ? mx(rc(p)) + rx(lc(p))
: rx(rc(p));

如果a[mid] == a[mid + 1],说明左右孩子不可以有交集,那答案就取两者最大:

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mx(p) = max(mx(lc(p)), mx(rc(p)));
lx(p) = lx(lc(p)), rx(p) = rx(rc(p));