제출 #765621

#제출 시각아이디문제언어결과실행 시간메모리
765621LucaIlieHomecoming (BOI18_homecoming)C++17
13 / 100
1083 ms51168 KiB
#include <bits/stdc++.h>
#include "homecoming.h"

using namespace std;

const long long INF = 1e16;
const int MAX_N = 2e6;

struct SegTree {
    vector<long long> segTree, lazy;

    void init( int n ) {
        segTree.resize( 4 * n );
        lazy.resize( 4 * n );
        for ( int v = 0; v < 4 * n; v++ ) {
            segTree[v] = -INF;
            lazy[v] = 0;
        }
    }

    void propag( int v, int l, int r ) {
        segTree[v] += lazy[v];
        if ( l != r ) {
            lazy[v * 2 + 1] += lazy[v];
            lazy[v * 2 + 2] += lazy[v];
        }
        lazy[v] = 0;
    }

    void update( int v, int l, int r, int lu, int ru, long long x ) {
        propag( v, l, r );

        if ( l > ru || r < lu )
            return;

        if ( lu <= l && r <= ru ) {
            lazy[v] = x;
            propag( v, l, r );
            return;
        }

        int mid = (l + r) / 2;
        update( v * 2 + 1, l, mid, lu, ru, x );
        update( v * 2 + 2, mid + 1, r, lu, ru, x );
        segTree[v] = max( segTree[v * 2 + 1], segTree[v * 2 + 2] );
    }

    long long query( int v, int l, int r, int lq, int rq ) {
        propag( v, l, r );

        if ( l > rq || r < lq )
            return -INF;

        if ( lq <= l && r <= rq )
            return segTree[v];

        int mid = (l + r) / 2;
        return max( query( v * 2 + 1, l, mid, lq, rq ), query( v * 2 + 2, mid + 1, r, lq, rq ) );
    }
};

SegTree dp;

long long solve( int n, int k, int a[], int b[] ) {
    long long ans = 0;

    for ( int l = 0; l <= k; l++ ) {
        dp.init( n + l + 1 );
        dp.update( 0, 0, n + l, n + l, n + l, INF );
        for ( int i = n - 1; i >= 0; i-- ) {
            if ( i >= l ) {
                long long dpii = 0;
                dpii = dp.query( 0, 0, n + l, i + 1, n + l ) - dp.query( 0, 0, n + l, i, i );
                dp.update( 0, 0, n + l, i, i, dpii );
            }

            dp.update( 0, 0, n + l, i + 1, n + l, -b[i] );
            dp.update( 0, 0, n + l, i + k, n + l, a[i] );
        }

        ans = max( ans, dp.query( 0, 0, n + l, 0, n + l ) );
    }

    return ans;
}
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