結果
| 問題 | No.3652 Range Bracket Sequence |
| コンテスト | |
| ユーザー |
|
| 提出日時 | 2026-08-28 22:21:52 |
| 言語 | C++23 (gcc 15.2.0 + boost 1.90.0) |
| 結果 |
AC
|
| 実行時間 | 113 ms / 2,000 ms |
| + 498µs | |
| コード長 | 5,341 bytes |
| 記録 | |
| コンパイル時間 | 2,308 ms |
| コンパイル使用メモリ | 338,812 KB |
| 実行使用メモリ | 16,128 KB |
| 最終ジャッジ日時 | 2026-08-28 22:22:04 |
| 合計ジャッジ時間 | 10,682 ms |
|
ジャッジサーバーID (参考情報) |
judge2_0 / judge3_0 |
(要ログイン)
| ファイルパターン | 結果 |
|---|---|
| sample | AC * 3 |
| other | AC * 57 |
ソースコード
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define ld long double
using LL = long long; using ULL = unsigned long long;
using VI = vector<int>; using VVI = vector<VI>; using VVVI = vector<VVI>;
using VL = vector<LL>; using VVL = vector<VL>; using VVVL = vector<VVL>;
using VB = vector<bool>; using VVB = vector<VB>; using VVVB = vector<VVB>;
using VD = vector<double>; using VVD = vector<VD>; using VVVD = vector<VVD>;
using VC = vector<char>; using VS = vector<string>; using VVC = vector<VC>;
using PII = pair<int,int>; using PLL = pair<LL,LL>; using PDD = pair<double,double>; using PIL = pair<int,LL>;
using MII = map<int,int>; using MLL = map<LL,LL>;
using SI = set<int>; using SL = set<LL>;
using MSI = multiset<int>; using MSL = multiset<LL>;
template<class T> using MAXPQ = priority_queue<T>;
template<class T> using MINPQ = priority_queue< T, vector<T>, greater<T> >;
const ll MOD = 1000000007;
const ll MOD2 = 998244353;
const ll INF = 1LL << 60;
#define PI 3.14159265358979323846
#define FOR(i, a, b) for(int i = (a); i < (b); ++i)
#define REP(i, n) FOR(i, 0, n)
#define EACH(e, v) for(auto &e : v)
#define RITR(it, v) for(auto it = (v).rbegin(); it != (v).rend(); ++it)
#define ALL(v) v.begin(),v.end()
vector<ll> x8={1,1,1,0,0,-1,-1,-1},y8={1,0,-1,1,-1,1,0,-1};
int dx4[4]={1,-1,0,0}, dy4[4]={0,0,1,-1};
/*
memo
-uf,RMQ(segtree),BIT,BIT2,SegTree,SegTreeLazy
-isprime,Eratosthenes,gcdlcm,factorize,divisors,modpow,moddiv
nCr(+modnCr,inverse,extend_euclid.powmod),tobaseB,tobase10
-dijkstra,Floyd,bellmanford,sccd,topological,treediamiter
-compress1,compress2,rotate90
-co,ci,fo1,fo2,fo3,fo4
-bitsearch,binaryserach
-bfs
-SegTreedec,SegTreeLazydec
*/
template <typename X>
struct SegTree{
using FX = function<X(X,X)>; //Xを2つ受け取りXを返す関数の型
long long n,N;
FX fx;
X ex;
vector<X> dat;
//要素数N,二項演算fx,単位元ex;
SegTree(long long _n, FX _fx, X _ex){
init(_n, _fx, _ex);
}
void init(long long _n, FX _fx, X _ex){
N = _n, fx = _fx; ex = _ex;
long long x = 1;
while(_n > x) x *= 2;
n = x;
dat.assign(n*2,_ex);
}
//i番目の要素にアクセス(0-indexed),O(1)
X operator[](long long i){return dat[n+i];}
X get(long long i){return dat[n+i];}
//i番目の要素をxにアップデート(0-indexed),O(logN)
void set(long long i, X x){
i += n;
dat[i] = x;
while(i >>= 1){
dat[i] = fx(dat[2*i],dat[2*i+1]);
}
}
//[l,r)で二項演算を作用した結果(0-indexed),O(logN)
X prod(long long l, long long r){
X vleft = ex, vright = ex;
for(long long left = l+n, right = r+n; left < right; left >>= 1, right >>= 1){
if(left & 1) vleft = fx(vleft,dat[left++]);
if(right & 1) vright = fx(dat[--right],vright);
}
return fx(vleft,vright);
}
//[0,N)まで二項演算を作用,O(1)
X all_prod() {return dat[1];}
//x=prod(l,r),f(x)=trueとなる最大のrを求める(f(ex)=true, 0-indexed),O(logN)
long long max_right(const function<bool(X)> f, long long l = 0){
if(l == N) return N;
l += n;
X sum = ex;
do{
while(l % 2 == 0) l >>= 1;
if(!f(fx(sum,dat[l]))){
while(l < n){
l = l * 2;
if(f(fx(sum,dat[l]))){
sum = fx(sum,dat[l]);
l++;
}
}
return l - n;
}
sum = fx(sum,dat[l]);
l++;
}while((l & -l) != l);
return N;
}
//x=prod(l,r),f(x)=trueとなる最小のlを求める(f(ex)=true, 0-indexed),O(logN)
long long min_left(const function<bool(X)> f, long long r = -1){
if(r == 0) return 0;
if(r == -1) return N;
r += n;
X sum = ex;
do{
r--;
while(r > 1 && (r % 2)) r >>= 1;
if(!f(fx(dat[r],sum))){
while(r < n){
r = r * 2 + 1;
if(f(fx(dat[r],sum))){
sum = fx(dat[r],sum);
r--;
}
}
return r + 1 - n;
}
sum = fx(dat[r],sum);
}while((r & -r) != r);
return 0;
}
};
struct dat{
ll c1=0,c2=0,match=0;
};
int main(){
cin.tie(0);
ios_base::sync_with_stdio(0);
ll N,Q; cin >> N >> Q;
string s; cin >> s;
auto fx = [&](dat l, dat r) -> dat{
ll p = min(l.c1-l.match,r.c2-r.match);
return dat{l.c1+r.c1,l.c2+r.c2,l.match+r.match+p};
};
dat ex = dat{0,0,0};
SegTree<dat> seg(N,fx,ex);
for(ll i = 0; i < N; i++){
if(s[i]=='('){
seg.set(i,dat{1,0,0});
}
else{
seg.set(i,dat{0,1,0});
}
}
while(Q--){
int t,a,b; cin >> t >> a >> b;
a--;
if(t==1){
if(b==1){
seg.set(a,dat{1,0,0});
}
else{
seg.set(a,dat{0,1,0});
}
}
if(t==2){
cout << seg.prod(a,b).match*2 << '\n';
}
}
}