제출 #550289

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
550289		BalintR	Team Contest (JOI22_team)	C++17	73 / 100	2047 ms	46056 KiB

team

#include <bits/stdc++.h>
using namespace std;
#pragma GCC optimize "Ofast"
#pragma GCC target "avx2,avx2"
 
typedef unsigned uint;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<pii> vpii;
typedef complex<double> cmplx;
template <typename T> using minPq = priority_queue<T, vector<T>, greater<T>>;
#define boost() cin.sync_with_stdio(0); cin.tie(0)
#define ms(a, x) memset(a, x, sizeof(a))
#define pb push_back
#define fs first
#define sn second
#define ALL(v) (v).begin(), (v).end()
#define SZ(v) ((int) (v).size())
#define lbv(v, x) (lower_bound((v).begin(), (v).end(), x) - (v).begin())
#define ubv(v, x) (upper_bound((v).begin(), (v).end(), x) - (v).begin())
template <typename T> inline void UNIQUE(vector<T> &v){sort(ALL(v)); v.resize(unique(ALL(v)) - v.begin());}
const int INF = 0x3f3f3f3f;
const ll LLINF = 0x3f3f3f3f3f3f3f3f;
const double PI = acos(-1);
#define FR(i, n) for(int i = 0; i < (n); i++)
#define FOR(i, a, b) for(int i = (a); i < (b); i++)
#define FORR(i, a, b) for(int i = (a); i >= (b); i--)
#define dbg(x) {cout << #x << ' ' << x << endl;}
#define dbgArr(arr, n) {cout << #arr; FR(_i, n) cout << ' ' << arr[_i]; cout << endl;}
 
template<typename T, int MSZ>
struct SegTree {
    T data[MSZ*2];
    int sz = MSZ;
    static constexpr T unit = 0;
 
    inline T merge(T t1, T t2){
        return max(t1, t2);
    }
 
    void build(){
        for (int i = sz-1; i > 0; --i){
            data[i] = merge(data[i*2], data[i*2+1]);
        }
    }
 
    void update(int p, T v){
        p += sz;
        data[p] = max(data[p], v);
        for (; p >>= 1;){
            data[p] = merge(data[p*2], data[p*2+1]);
        }
    }
 
    // Query range [l, r)
    T query(int l, int r){
        T resL = unit, resR = unit;
        for (l += sz, r += sz; l < r; l >>= 1, r >>= 1) {
            if(l&1) resL = merge(resL, data[l++]);
            if(r&1) resR = merge(data[--r], resR);
        }
        return merge(resL, resR);
    }
 
    // Access the last layer of the segment tree
    inline T& operator[](int i){
        return data[i + sz];
    }
};
 
typedef array<int, 3> trip;
 
const int MN = 150005;
int n;
vector<trip> vec;
vi ys, zs;
SegTree<int, MN> mxZPref, mxYPref;
int mxZ[MN];
 
int main(){
    boost();
    cin >> n;
    FR(i, n){
        int a, b, c;
        cin >> a >> b >> c;
        vec.pb({a, b, c});
        ys.pb(b);
        zs.pb(c);
    }
 
    UNIQUE(vec);
    UNIQUE(ys);
    UNIQUE(zs);
    n = SZ(ys);
 
    int ans = -1;
    int lst = 0, prvX = vec[0][0];
 
    FR(i, SZ(vec)){
        if(vec[i][0] != prvX){
            FOR(j, lst, i){
                auto [x, y, z] = vec[j];
                int yPos = lbv(ys, y);
                int zPos = lbv(zs, z);
 
                int mz = mxZPref.query(0, yPos);
                if(mz > z) mxZ[yPos] = max(mxZ[yPos], mz);
                mxYPref.update(zPos, y);
 
                int mxY = mxYPref.query(0, zPos);
                if(mxY > y){
                    int yp1 = lbv(ys, mxY);
                    mxZ[yp1] = max(mxZ[yp1], z);
                }
                mxZPref.update(yPos, z);
            }
 
            prvX = vec[i][0];
            lst = i;
        }
 
        auto [x, y, z] = vec[i];
        int yPos = lbv(ys, y);
        int mxSum = 0;
 
        FOR(j, yPos+1, n) mxSum = max(mxSum, (ys[j] + mxZ[j]) * (mxZ[j] > z));
        if(mxSum) ans = max(ans, mxSum + x);
    }
 
    if(ans == -1) vi vec(1e7, 1234);
 
    cout << ans << '\n';
}

• Subtask 1: 8 / 8
• Subtask 2: 29 / 29
• Subtask 3: 9 / 9
• Subtask 4: 9 / 9
• Subtask 5: 9 / 9
• Subtask 6: 9 / 9
• Subtask 7: 0 / 27