結果

問題 No.776 A Simple RMQ Problem
ユーザー fumiphys
提出日時 2019-03-12 23:36:16
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 479 ms / 3,000 ms
コード長 4,466 bytes
コンパイル時間 1,322 ms
コンパイル使用メモリ 125,748 KB
実行使用メモリ 18,508 KB
最終ジャッジ日時 2024-06-23 15:54:55
合計ジャッジ時間 11,466 ms
ジャッジサーバーID
(参考情報)
judge2 / judge5
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 2
other AC * 26
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

// includes
#include <cstdio>
#include <cstdint>
#include <iostream>
#include <iomanip>
#include <string>
#include <queue>
#include <stack>
#include <vector>
#include <set>
#include <map>
#include <unordered_map>
#include <algorithm>
#include <utility>
#include <functional>
#include <cmath>
#include <climits>
#include <bitset>
#include <list>
#include <random>
// macros
#define ll long long int
#define pb emplace_back
#define mk make_pair
#define pq priority_queue
#define FOR(i, a, b) for(int i=(a);i<(b);++i)
#define rep(i, n) FOR(i, 0, n)
#define rrep(i, n) for(int i=((int)(n)-1);i>=0;i--)
#define all(x) (x).begin(),(x).end()
#define sz(x) ((int)(x).size())
#define UNIQUE(v) v.erase(unique(v.begin(), v.end()), v.end())
using namespace std;
// types
typedef pair<int, int> P;
typedef pair<ll, int> Pl;
typedef pair<ll, ll> Pll;
typedef pair<double, double> Pd;
// constants
const int inf = 1e9;
const ll linf = 1LL << 50;
const double EPS = 1e-10;
const int mod = 1e9 + 7;
// solve
template <class T>bool chmax(T &a, const T &b){if(a < b){a = b; return 1;} return 0;}
template <class T>bool chmin(T &a, const T &b){if(a > b){a = b; return 1;} return 0;}
template<typename T, typename E>
struct SegmentTree_ {
function<T(T, T)> f;
function<T(T, E)> g;
int n;
T def;
vector<T> vec;
SegmentTree_(){}
SegmentTree_(int n_, function<T(T, T)> f_, function<T(T, E)> g_, T def_, vector<T> v=vector<T>()){
f = f_;
g = g_;
def = def_;
// initialize vector
n = 1;
while(n < n_){
n *= 2;
}
vec = vector<T>(2*n -1, def);
// initialize segment tree
for(int i = 0; i < v.size(); i++){
vec[i + n - 1] = v[i];
}
for(int i = n - 2; i >= 0; i--){
vec[i] = f(vec[2*i+1], vec[2*i+2]);
}
}
void update(int k, E val){
k = k + n - 1;
vec[k] = g(vec[k], val);
while(k > 0){
k = (k - 1) / 2;
vec[k] = f(vec[2*k+1], vec[2*k+2]);
}
}
// [l, r) -> [a, b) (at k)
T query(int a, int b, int k, int l, int r){
if(r <= a || b <= l)return def;
if(a <= l && r <= b)return vec[k];
T ld = query(a, b, 2*k+1, l, (l+r)/2);
T rd = query(a, b, 2*k+2, (l+r)/2, r);
return f(ld, rd);
}
T query(int a, int b){
return query(a, b, 0, 0, n);
}
};
template<typename T, typename E>
using SegmentTree = struct SegmentTree_<T, E>;
using SegmentTreeI = SegmentTree<int, int>;
struct section{
ll sum, lsum, rsum, msum;
section(): section(0, -linf, -linf, -linf){}
section(ll sum, ll lsum, ll rsum, ll msum): sum(sum), lsum(lsum), rsum(rsum), msum(msum){}
section operator+(const section &right) const{
section res;
res.sum = sum + right.sum;
res.lsum = max(lsum, sum + right.lsum);
res.rsum = max(right.rsum, right.sum + rsum);
res.msum = max(msum, max(right.msum, rsum + right.lsum));
return res;
}
};
int main(int argc, char const* argv[])
{
ios_base::sync_with_stdio(false);
cin.tie(0);
int n, q;
cin >> n >> q;
vector<ll> a(n);
rep(i, n)cin >> a[i];
vector<section> vec(n);
for(int i = 0; i < n; i++){
vec[i] = section(a[i], a[i], a[i], a[i]);
}
SegmentTree<section, section> seg = SegmentTree<section, section>(n, [](section a, section b){return a + b;},
[](section a, section b){return b;}, section(), vec);
rep(i_, q){
string s;
cin >> s;
if(s == "set"){
int i;
ll x;
cin >> i >> x;
i--;
seg.update(i, section(x, x, x, x));
}else{
int l1, l2, r1, r2;
cin >> l1 >> l2 >> r1 >> r2;
l1--, l2--, r1--, r2--;
ll res = 0;
if(l2 < r1){
res += seg.query(l1, l2+1).rsum;
res += seg.query(l2+1, r1).sum;
res += seg.query(r1, r2+1).lsum;
}else{
r1 = max(r1, l1);
l2 = min(l2, r2);
res = -linf;
ll tmp = 0;
tmp += seg.query(l1, r1).rsum;
tmp += seg.query(r1, l2+1).sum;
tmp += seg.query(l2+1, r2+1).lsum;
res = max(res, tmp);
tmp = 0;
tmp += seg.query(r1, l2+1).rsum;
tmp += seg.query(l2+1, r2+1).lsum;
res = max(res, tmp);
tmp = 0;
tmp += seg.query(l1, r1).rsum;
tmp += seg.query(r1, l2+1).lsum;
res = max(res, tmp);
tmp = 0;
tmp += seg.query(r1, l2+1).msum;
res = max(res, tmp);
}
cout << res << endl;
}
//for(int i = 0; i < n; i++)cout << seg.query(i, i+1).rsum << "\n "[i + 1 != n];
}
return 0;
}
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