P3870 开关

考点

  • 线段树

题解

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#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 50;
int n, m;

// s:1的个数总和
struct {
int l, r, s, tag;
#define l(x) tr[x].l
#define r(x) tr[x].r
#define s(x) tr[x].s
#define tag(x) tr[x].tag
#define lc(x) x * 2
#define rc(x) x * 2 + 1
} tr[4 * maxn];

void up(int p) { s(p) = s(lc(p)) + s(rc(p)); }

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

void down(int p) {
if (tag(p)) {
s(lc(p)) = r(lc(p)) - l(lc(p)) + 1 - s(lc(p));
s(rc(p)) = r(rc(p)) - l(rc(p)) + 1 - s(rc(p));
tag(lc(p)) ^= 1, tag(rc(p)) ^= 1, tag(p) ^= 1;
}
}

void update(int p, int l, int r) {
if (l(p) >= l && r(p) <= r) {
s(p) = r(p) - l(p) + 1 - s(p);
tag(p) ^= 1;
return;
}
down(p);
int mid = (l(p) + r(p)) / 2;
if (l <= mid) update(lc(p), l, r);
if (r > mid) update(rc(p), l, r);
up(p);
}

int ask(int p, int l, int r) {
if (l(p) >= l && r(p) <= r) return s(p);
down(p);
int mid = (l(p) + r(p)) / 2;
int ans = 0;
if (l <= mid) ans += ask(lc(p), l, r);
if (r > mid) ans += ask(rc(p), l, r);
return ans;
}

int main() {
scanf("%d%d", &n, &m);
int opt, x, y;
build(1, 1, n);
while (m--) {
scanf("%d%d%d", &opt, &x, &y);
if (!opt) {
update(1, x, y);
} else {
printf("%d\n", ask(1, x, y));
}
}
return 0;
}

思路

假设当前区间为[1, 4]1的个数s3,那么0的个数等于4 - 3 = 1

如果对该区间取反,那么1的个数s等于4 - 3 = 10的个数等于4 - 1 = 3

延迟标记tag应该使用异或,因为取反的取反就是保持原样。