[Codeforces 915E]Physical Education Lessons

Description

题库链接

给你一个长度为 (n) 的 (01) 序列。初始全为 (1) ，让你兹磁 (m) 次操作：将区间内的各个数取反，并询问整个区间的 (1) 的个数。

(1leq nleq 10^9,1leq mleq 3cdot 10^5)

Solution

可以将所有询问离线，对于序列上端点间的一整块直接缩成一个节点。点权为区间长度，点数是 (O(m)) 的，线段树维护就好了。

然鹅，大佬都是直接动态开点做的...

Code

#include <bits/stdc++.h>
using namespace std;
const int N = 3e5;

struct operate {int l, r, id; }a[N+5];
int n, m, b[N*2+5], id[N*2+5], cost[N*4+5], tot, pos;
struct Segment_tree {
#define lr(o) (o<<1)
#define rr(o) (o<<1|1)
  int sgm[N*16+5], sum[N*16+5], lazy[N*16+5];
  void build(int o, int l, int r) {
    if (l == r) {sgm[o] = sum[o] = cost[l]; return; }
    int mid = (l+r)>>1;
    build(lr(o), l, mid); build(rr(o), mid+1, r);
    sum[o] = sum[lr(o)]+sum[rr(o)];
    sgm[o] = sgm[lr(o)]+sgm[rr(o)];
  }
  void pushdown(int o) {
    if (lazy[o] == -1) sgm[lr(o)] = sgm[rr(o)] = 0, lazy[lr(o)] = lazy[rr(o)] = -1;
    else sgm[lr(o)] = sum[lr(o)], sgm[rr(o)] = sum[rr(o)], lazy[lr(o)] = lazy[rr(o)] = 1;
    lazy[o] = 0;
  }
  void update(int o, int l, int r, int a, int b, int val) {
    if (a <= l && r <= b) {
      if (val == 1) sgm[o] = 0, lazy[o] = -1;
      else sgm[o] = sum[o], lazy[o] = 1;
      return;
    }
    if (lazy[o]) pushdown(o); int mid = (l+r)>>1;
    if (a <= mid) update(lr(o), l, mid, a, b, val);
    if (b > mid) update(rr(o), mid+1, r, a, b, val);
    sgm[o] = sgm[lr(o)]+sgm[rr(o)];
  }
}T;

void work() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d%d", &a[i].l, &a[i].r, &a[i].id);
    b[++tot] = a[i].l, b[++tot] = a[i].r;
  }
  sort(b+1, b+tot+1); tot = unique(b+1, b+tot+1)-b-1;
  if (b[1] != 1) ++pos, cost[pos] = b[1]-1;
  for (int i = 1; i < tot; i++) {
    id[i] = ++pos, cost[pos] = 1;
    if (b[i]+1 != b[i+1]) ++pos, cost[pos] = b[i+1]-b[i]-1;
  }
  id[tot] = ++pos, cost[pos] = 1;
  if (b[tot] != n) ++pos, cost[pos] = n-b[tot];
  T.build(1, 1, pos);
  for (int i = 1; i <= m; i++) {
    int l = id[lower_bound(b+1, b+tot+1, a[i].l)-b], r = id[lower_bound(b+1, b+tot+1, a[i].r)-b];
    T.update(1, 1, pos, l, r, a[i].id);
    printf("%d
", T.sgm[1]);
  }
}
int main() {work(); return 0; }