Description
给定 01 01 01 序列 a = ( a 1 , a 2 , ⋯ , a n ) a=(a_1,a_2,\cdots,a_n) a=(a1,a2,⋯,an)，并定义 f ( l , r ) = [ ( ∑ i = l r a i ) = r − l + 1 ] f(l,r)=[(\sum\limits_{i=l}^r a_i)=r-l+1] f(l,r)=[(i=l∑rai)=r−l+1].
执行 m m m 个操作，分五种：
- reset ( l , r ) \operatorname{reset}(l,r) reset(l,r)：对每个 i ∈ [ l , r ] i\in[l,r] i∈[l,r] 执行 a i ← 0 a_i\gets 0 ai←0.
- set ( l , r ) \operatorname{set}(l,r) set(l,r)：对每个 i ∈ [ l , r ] i\in[l,r] i∈[l,r] 执行 a i ← 1 a_i\gets 1 ai←1.
- negate ( l , r ) \operatorname{negate}(l,r) negate(l,r)：对每个 i ∈ [ l , r ] i\in[l,r] i∈[l,r] 执行 a i ← 1 − a i a_i\gets 1-a_i ai←1−ai.
- sum ( l , r ) \operatorname{sum}(l,r) sum(l,r)：求 ∑ i = l r a i \sum\limits_{i=l}^r a_i i=l∑rai.
- gss ( l , r ) \operatorname{gss}(l,r) gss(l,r)：求 max [ s , t ] ∈ [ l , r ] ( t − s + 1 ) × f ( s , t ) \max\limits_{[s,t]\in[l,r]}(t-s+1)\times f(s,t) [s,t]∈[l,r]max(t−s+1)×f(s,t).
Limitations
1 ≤ n , m ≤ 1 0 5 1\le n,m\le 10^5 1≤n,m≤105
1 ≤ l ≤ r ≤ n 1\le l\le r\le n 1≤l≤r≤n
1 s , 128 MB 1\text{s},128\text{MB} 1s,128MB
Solution
线段树做法.
节点上维护 S = ( sum , lsum , rsum , tsum , len ) S=(\textit{sum},\textit{lsum},\textit{rsum},\textit{tsum},\textit{len}) S=(sum,lsum,rsum,tsum,len)，用类似最大子段和的方法更新.
由于要取反， 0 0 0 和 1 1 1 都分别需要一个 S S S（下记为 S 0 S_0 S0 和 S 1 S_1 S1） .
然后需要标记 T T T，显然 T = ( cov , rev ) T=(\textit{cov},\textit{rev}) T=(cov,rev)（分别为赋值和取反标记）
- 考虑赋值，我们直接将 S c o v S_{cov} Scov 全部填上 len \textit{len} len， S 1 − c o v S_{1-cov} S1−cov 全部清零（ len \textit{len} len 要不变）
- 考虑翻转，我们直接交换 S 0 S_0 S0 和 S 1 S_1 S1.
然后需要考虑 T + T T+T T+T：
- 赋值可以直接打标记.
- 对于取反，当一个区间赋值后，取反就相当于赋值 ( 1 − cov ) (1-\textit{cov}) (1−cov)，那么我们可以将 cov \textit{cov} cov 取反（此时 rev \textit{rev} rev 没用，要置为 0 0 0），否则，直接更新 rev \textit{rev} rev.
有不少坑点：
- 下标从 0 0 0 开始.
- 没有被赋值的节点， cov \textit{cov} cov 要置为 − 1 -1 −1 表示没有赋值.
- 更新 S S S 时，左半区间满了， lsum \textit{lsum} lsum 才能跨越中点， rsum \textit{rsum} rsum 同理.
- 先处理 cov \textit{cov} cov 再处理 rev \textit{rev} rev.
- 打 rev \textit{rev} rev 标记时
^= 1不要写成= 1.
Code
3.9 KB , 0.45 s , 11.63 MB (in total, C++20 with O2) 3.9\text{KB},0.45\text{s},11.63\text{MB}\;\texttt{(in total, C++20 with O2)} 3.9KB,0.45s,11.63MB(in total, C++20 with O2)
建议封装 S S S（见代码中的 Data），会好写不少.
cpp
#include <bits/stdc++.h>
using namespace std;
using i64 = long long;
using ui64 = unsigned long long;
using i128 = __int128;
using ui128 = unsigned __int128;
using f4 = float;
using f8 = double;
using f16 = long double;
template<class T>
bool chmax(T &a, const T &b){
if(a < b){ a = b; return true; }
return false;
}
template<class T>
bool chmin(T &a, const T &b){
if(a > b){ a = b; return true; }
return false;
}
namespace seg_tree {
struct Data {
int sum, lsum, rsum, tsum, len;
inline Data() {}
inline Data(int x) : sum(x), lsum(x), rsum(x), tsum(x), len(1) {}
inline Data(int _sum, int _lsum, int _rsum, int _tsum, int _len)
: sum(_sum), lsum(_lsum), rsum(_rsum), tsum(_tsum), len(_len) {}
};
inline Data operator+(const Data& lhs, const Data& rhs) {
const int _sum = lhs.sum + rhs.sum;
const int _lsum = lhs.lsum + (lhs.sum == lhs.len) * rhs.lsum;
const int _rsum = rhs.rsum + (rhs.sum == rhs.len) * lhs.rsum;
const int _tsum = std::max({lhs.tsum, rhs.tsum, lhs.rsum + rhs.lsum});
const int _len = lhs.len + rhs.len;
return Data(_sum, _lsum, _rsum, _tsum, _len);
}
struct Node {
int l, r;
array<Data, 2> dat;
int cov, rev;
inline Data& operator[](int i) { return dat[i]; }
inline Data operator[](int i) const { return dat[i]; }
};
inline int ls(int u) { return 2 * u + 1; }
inline int rs(int u) { return 2 * u + 2; }
struct SegTree {
vector<Node> tr;
inline SegTree() {}
inline SegTree(const vector<int>& a) {
const int n = a.size();
tr.resize(n << 1);
build(0, 0, n - 1, a);
}
inline void pushup(int u, int mid) {
for (int i = 0; i < 2; i++) tr[u][i] = tr[ls(mid)][i] + tr[rs(mid)][i];
}
inline void apply(int u, int cov, int rev) {
if (~cov) {
const int len = tr[u][cov].len;
tr[u][cov] = Data(len, len, len, len, len);
tr[u][cov ^ 1] = Data(0, 0, 0, 0, len);
tr[u].cov = cov;
tr[u].rev = 0;
return;
}
if (rev) {
swap(tr[u][0], tr[u][1]);
if (~tr[u].cov) tr[u].cov ^= 1;
else tr[u].rev ^= 1;
}
}
inline void pushdown(int u, int mid) {
apply(ls(mid), tr[u].cov, tr[u].rev);
apply(rs(mid), tr[u].cov, tr[u].rev);
tr[u].cov = -1, tr[u].rev = 0;
}
void build(int u, int l, int r, const vector<int>& a) {
tr[u].l = l, tr[u].r = r, tr[u].cov = -1;
if (l == r) {
for (int i = 0; i < 2; i++) tr[u][i] = Data(a[l] == i);
return;
}
const int mid = (l + r) >> 1;
build(ls(mid), l, mid, a);
build(rs(mid), mid + 1, r, a);
pushup(u, mid);
}
void update(int u, int l, int r, int cov, int rev) {
if (l <= tr[u].l && tr[u].r <= r) return apply(u, cov, rev);
const int mid = (tr[u].l + tr[u].r) >> 1;
pushdown(u, mid);
if (l <= mid) update(ls(mid), l, r, cov, rev);
if (r > mid) update(rs(mid), l, r, cov, rev);
pushup(u, mid);
}
Data query(int u, int l, int r) {
if (l <= tr[u].l && tr[u].r <= r) return tr[u][1];
const int mid = (tr[u].l + tr[u].r) >> 1;
pushdown(u, mid);
if (r <= mid) return query(ls(mid), l, r);
else if (l > mid) return query(rs(mid), l, r);
else return query(ls(mid), l, r) + query(rs(mid), l, r);
}
inline void range_cover(int l, int r, int v) { update(0, l, r, v, 0); }
inline void range_negate(int l, int r) { update(0, l, r, -1, 1); }
inline int range_sum(int l, int r) { return query(0, l, r).sum; }
inline int range_gss(int l, int r) { return query(0, l, r).tsum; }
};
}
using seg_tree::SegTree;
signed main() {
ios::sync_with_stdio(0);
cin.tie(0), cout.tie(0);
int n, m;
scanf("%d %d", &n, &m);
vector<int> a(n);
for (int i = 0; i < n; i++) scanf("%d", &a[i]);
SegTree sgt(a);
for (int i = 0, op, l, r; i < m; i++) {
scanf("%d %d %d", &op, &l, &r);
if (op == 0) sgt.range_cover(l, r, 0);
if (op == 1) sgt.range_cover(l, r, 1);
if (op == 2) sgt.range_negate(l, r);
if (op == 3) printf("%d\n", sgt.range_sum(l, r));
if (op == 4) printf("%d\n", sgt.range_gss(l, r));
}
return 0;
}