Submission #767098

#	Time	Username	Problem	Language	Result	Execution time	Memory
767098		LucaIlie	Homecoming (BOI18_homecoming)	C++17	31 / 100	1041 ms	169128 KiB

homecoming

#include <bits/stdc++.h>
#include "homecoming.h"

using namespace std;

const long long INF = 1e16;
const int MAX_SGT = 4e6 + 1;

struct SegTree {
    long long segTree[4 * MAX_SGT], lazy[4 * MAX_SGT];

    void init( int 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 sumB[MAX_SGT], val[MAX_SGT];

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

    for ( int i = n; i <= n + k; i++ )
        sumB[i] = 0;
    for ( int i = n - 1; i >= 0; i-- )
        sumB[i] = sumB[i + 1] + b[i];

    for ( int take = 0; take <= 1; take++ ) {
        dp.init( n + k + 1 );
        if ( !take ) {
            long long s = 0;
            for ( int j = 0; j <= k; j++ ) {
                val[n + j] = -s;
                s += b[j];
            }
        } else {
            for ( int j = 0; j <= k; j++ )
                val[n + j] = 0;
        }
        for ( int j = 0; j <= k; j++ )
            dp.update( 0, 0, n + k, n + j, n + j, INF + val[n + j] );


        long long maxA = val[n + k];
        for ( int i = n - 1; i >= 0; i-- ) {
            val[i] = -INF;
            if ( i >= k * take ) {
                val[i] = max( dp.query( 0, 0, n + k, i + 1, i + k ), maxA );
                dp.update( 0, 0, n + k, i, i, INF + val[i] );
            }

            dp.update( 0, 0, n + k, i + 1, i + k - 1, -b[i] );
            maxA = max( maxA, val[i + k] - (sumB[i + 1] - sumB[i + k]) );
            maxA = maxA - b[i] + a[i];
        }

        ans = max( ans, dp.query( 0, 0, n + k, 0, k - 1 ) );
        ans = max( ans, maxA );
    }

    return ans;
}

• Subtask 1: 13 / 13
• Subtask 2: 18 / 18
• Subtask 3: 0 / 31
• Subtask 4: 0 / 38