题目链接🔗：2643. 序列操作 - AcWing题库

前驱知识：需要理解线段树的结构和程序基本框架、以及懒标记的操作。

题目描述

题目分析 

对区间在线进行修改和查询，一般就是用线段树来解决，观察到题目一共有五个操作：

我们首先要思考需要用线段树维护哪些信息，通过维护这些信息，在查询时能够得到需要的答案。

根据查询的内容，我们发现需要维护区间内1的个数，以及区间内最多连续1的个数

由于题目的操作对象就是0和1，我们可以想到对称维护0和1的信息（主要是因为存在操作2

那么综合来看，我们可以维护线段树的以下信息：

 ：区间左端点

 ：区间右端点

 ：区间内1的个数

 ：区间内0的个数

 ：区间内左边最多连续1的个数

 ：区间内左边最多连续0的个数

 ：区间内右边最多连续1的个数

 ：区间内右边最多连续0的个数

 ：区间内最长连续0的个数

：区间内最长连续1的个数

 ：操作0对应的懒标记

 ：操作1对应的懒标记

 ：操作2对应的懒标记

---

具体维护方案如下

AC代码 

#include 
#include 
#include using namespace std ;const int N = 1e5 + 10 ; int n, m, a[N] ; 
struct Node 
{int l, r ; int sum0, sum1, l0, l1, r0, r1, m0, m1 ; bool flag0, flag1, rev ; 
}tr[4 * N] ; void pushup(int u) 
{tr[u].sum0 = tr[u << 1].sum0 + tr[u << 1 | 1].sum0 ; tr[u].sum1 = tr[u << 1].sum1 + tr[u << 1 | 1].sum1 ;tr[u].l0 = (tr[u << 1].sum1) ? tr[u << 1].l0 : tr[u << 1].sum0 + tr[u << 1 | 1].l0 ;tr[u].l1 = (tr[u << 1].sum0) ? tr[u << 1].l1 : tr[u << 1].sum1 + tr[u << 1 | 1].l1 ;tr[u].r0 = (tr[u << 1 | 1].sum1) ? tr[u << 1 | 1].r0 :  tr[u << 1 | 1].sum0 + tr[u << 1].r0 ; tr[u].r1 = (tr[u << 1 | 1].sum0) ? tr[u << 1 | 1].r1 :  tr[u << 1 | 1].sum1 + tr[u << 1].r1 ;tr[u].m0 = max(max(tr[u << 1].m0, tr[u << 1 | 1].m0), tr[u << 1].r0 + tr[u << 1 | 1].l0) ;tr[u].m1 = max(max(tr[u << 1].m1, tr[u << 1 | 1].m1), tr[u << 1].r1 + tr[u << 1 | 1].l1) ;
}void pushup_Node(Node &root, Node ls, Node rs) 
{root.sum0 = ls.sum0 + rs.sum0 ; root.sum1 = ls.sum1 + rs.sum1 ; root.l0 = (ls.sum1) ? ls.l0 : ls.sum0 + rs.l0 ; root.l1 = (ls.sum0) ? ls.l1 : ls.sum1 + rs.l1 ; root.r0 = (rs.sum1) ? rs.r0 : rs.sum0 + ls.r0 ; root.r1 = (rs.sum0) ? rs.r1 : rs.sum1 + ls.r1 ;root.m0 = max(max(ls.m0, rs.m0), ls.r0 + rs.l0) ; root.m1 = max(max(ls.m1, rs.m1), ls.r1 + rs.l1) ;
}void pushdown(int u) 
{if (tr[u].flag0) {tr[u << 1].sum0 = tr[u << 1].l0 = tr[u << 1].r0 = tr[u << 1].m0 = tr[u << 1].r - tr[u << 1].l + 1 ; tr[u << 1 | 1].sum0 = tr[u << 1 | 1].l0 = tr[u << 1 | 1].r0 = tr[u << 1 | 1].m0 = tr[u << 1 | 1].r - tr[u << 1 | 1].l + 1 ;tr[u << 1].sum1 = tr[u << 1].l1 = tr[u << 1].r1 = tr[u << 1].m1 = 0 ; tr[u << 1 | 1].sum1 = tr[u << 1 | 1].l1 = tr[u << 1 | 1].r1 = tr[u << 1 | 1].m1 = 0 ; tr[u << 1].flag0 = tr[u << 1 | 1].flag0 = true ; tr[u << 1].flag1 = tr[u << 1 | 1].flag1 = tr[u << 1].rev = tr[u << 1 | 1].rev = false ;tr[u].flag0 = false ; }if (tr[u].flag1) {tr[u << 1].sum1 = tr[u << 1].l1 = tr[u << 1].r1 = tr[u << 1].m1 = tr[u << 1].r - tr[u << 1].l + 1 ; tr[u << 1 | 1].sum1 = tr[u << 1 | 1].l1 = tr[u << 1 | 1].r1 = tr[u << 1 | 1].m1 = tr[u << 1 | 1].r - tr[u << 1 | 1].l + 1 ;tr[u << 1].sum0 = tr[u << 1].l0 = tr[u << 1].r0 = tr[u << 1].m0 = 0 ;tr[u << 1 | 1].sum0 = tr[u << 1 | 1].l0 = tr[u << 1 | 1].r0 = tr[u << 1 | 1].m0 = 0 ;tr[u << 1].flag1 = tr[u << 1 | 1].flag1 = true ;tr[u << 1 | 1].flag0 = tr[u << 1 | 1].flag0 = tr[u << 1].rev = tr[u << 1 | 1].rev = false ;tr[u].flag1 = false ;}if (tr[u].rev) {swap(tr[u << 1].sum0, tr[u << 1].sum1) ;swap(tr[u << 1 | 1].sum0, tr[u << 1 | 1].sum1) ;swap(tr[u << 1].l0, tr[u << 1].l1) ; swap(tr[u << 1 | 1].l0, tr[u << 1 | 1].l1) ;swap(tr[u << 1].r0, tr[u << 1].r1) ; swap(tr[u << 1 | 1].r0, tr[u << 1 | 1].r1) ;swap(tr[u << 1].m0, tr[u << 1].m1) ; swap(tr[u << 1 | 1].m0, tr[u << 1 | 1].m1) ;tr[u << 1].rev ^= 1, tr[u << 1 | 1].rev ^= 1 ; tr[u].rev = 0 ;}
}void build(int u, int l, int r) 
{tr[u].l = l, tr[u].r = r ; if (l == r) {tr[u].sum0 = tr[u].l0 = tr[u].r0 = tr[u].m0 = a[r] ^ 1 ; tr[u].sum1 = tr[u].l1 = tr[u].r1 = tr[u].m1 = a[r] & 1 ; return ; }int mid = l + r >> 1 ;build(u << 1, l, mid), build(u << 1 | 1, mid + 1, r) ; pushup(u) ; 
}void change1(int u, int l, int r) 
{if (tr[u].l >= l && tr[u].r <= r) {tr[u].sum1 = tr[u].l1 = tr[u].r1 = tr[u].m1 = 0 ; tr[u].sum0 = tr[u].l0 = tr[u].r0 = tr[u].m0 = tr[u].r - tr[u].l + 1 ; tr[u].flag0 = true, tr[u].flag1 = tr[u].rev = false ;return ;}pushdown(u) ; int mid = tr[u].l + tr[u].r >> 1 ; if (l <= mid) change1(u << 1, l, r) ; if (r > mid) change1(u << 1 | 1, l, r) ; pushup(u) ;
}void change2(int u, int l, int r) 
{if (tr[u].l >= l && tr[u].r <= r) {tr[u].sum0 = tr[u].l0 = tr[u].r0 = tr[u].m0 = 0 ; tr[u].sum1 = tr[u].l1 = tr[u].r1 = tr[u].m1 = tr[u].r - tr[u].l + 1 ; tr[u].flag1 = true, tr[u].flag0 = tr[u].rev = false ;return ;}pushdown(u) ; int mid = tr[u].l + tr[u].r >> 1 ; if (l <= mid) change2(u << 1, l, r) ; if (r > mid) change2(u << 1 | 1, l, r) ; pushup(u) ; 
}void Reverse(int u, int l, int r) 
{if (tr[u].l >= l && tr[u].r <= r) {swap(tr[u].sum0, tr[u].sum1) ; swap(tr[u].l0, tr[u].l1) ; swap(tr[u].r0, tr[u].r1) ; swap(tr[u].m0, tr[u].m1) ; tr[u].rev ^= 1 ; return ; }pushdown(u) ; int mid = tr[u].l + tr[u].r >> 1 ; if (l <= mid) Reverse(u << 1, l, r) ; if (r > mid) Reverse(u << 1 | 1, l, r) ;pushup(u) ; 
}int ask1(int u, int l, int r) 
{if (tr[u].l >= l && tr[u].r <= r) return tr[u].sum1 ; pushdown(u) ; int mid = tr[u].l + tr[u].r >> 1 ; int sum = 0 ; if (l <= mid) sum = ask1(u << 1, l, r) ; if (r > mid) sum += ask1(u << 1 | 1, l, r) ; return sum ;
}Node ask2(int u, int l, int r) 
{if (tr[u].l >= l && tr[u].r <= r) return tr[u] ; pushdown(u) ; int mid = tr[u].l + tr[u].r >> 1 ; Node res ; if (l > mid) return ask2(u << 1 | 1, l, r) ; if (r <= mid) return ask2(u << 1, l, r) ; Node ls = ask2(u << 1, l, r), rs = ask2(u << 1 | 1, l, r) ; pushup_Node(res, ls, rs) ; return res ; 
}int main() 
{ios::sync_with_stdio(false) ; cin >> n >> m ; for (int i = 1 ; i <= n ; i ++ ) cin >> a[i] ;build(1, 1, n) ; while (m -- ) {int opt, l, r ; cin >> opt >> l >> r ; l ++, r ++ ; if (opt == 0) change1(1, l, r) ; else if (opt == 1) change2(1, l, r) ; else if (opt == 2) Reverse(1, l, r) ; else if (opt == 3) cout << ask1(1, l , r) << endl ;else cout << ask2(1, l, r).m1 << endl ; }return 0 ; 
}