| # | Time | Username | Problem | Language | Result | Execution time | Memory | 
|---|---|---|---|---|---|---|---|
| 527823 | mathking1021 | Salesman (IOI09_salesman) | C++17 | 417 ms | 37348 KiB | 
This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include <iostream>
#include <cstdio>
#include <algorithm>
#define PLPLL pair<ll, PLL>
#define PLL pair<ll, ll>
#define F first
#define S second
using namespace std;
typedef long long ll;
const ll M = 1e18;
ll n, m, m1, m2, k;
PLPLL a[500005];
ll base = (1 << 19);
ll seg1[1100005];
ll seg2[1100005];
void update1(ll p, ll q)
{
    p += base;
    seg1[p] = q;
    p /= 2;
    while(p >= 1)
    {
        seg1[p] = max(seg1[p * 2], seg1[p * 2 + 1]);
        p /= 2;
    }
}
void update2(ll p, ll q)
{
    p += base;
    seg2[p] = q;
    p /= 2;
    while(p >= 1)
    {
        seg2[p] = max(seg2[p * 2], seg2[p * 2 + 1]);
        p /= 2;
    }
}
ll query1(ll p, ll q)
{
    p += base;
    q += base;
    ll re = -M;
    while(p < q)
    {
        if(p % 2 == 1) re = max(re, seg1[p]), p++;
        if(q % 2 == 0) re = max(re, seg1[q]), q--;
        p /= 2, q /= 2;
    }
    if(p == q) re = max(re, seg1[p]);
    return re;
}
ll query2(ll p, ll q)
{
    p += base;
    q += base;
    ll re = -M;
    while(p < q)
    {
        if(p % 2 == 1) re = max(re, seg2[p]), p++;
        if(q % 2 == 0) re = max(re, seg2[q]), q--;
        p /= 2, q /= 2;
    }
    if(p == q) re = max(re, seg2[p]);
    return re;
}
int main()
{
    scanf("%lld%lld%lld%lld", &n, &m1, &m2, &k);
    m = m1 + m2;
    for(ll i = 0; i < n; i++)
    {
        scanf("%lld%lld%lld", &a[i].F, &a[i].S.F, &a[i].S.S);
    }
    sort(a, a + n);
    for(ll i = 1; i < 2 * base; i++) seg1[i] = seg2[i] = -M;
    update1(k, 0 - k * m);
    update2(k, 0);
    for(ll i = 0; i < n; i++)
    {
        ll t1 = query1(a[i].S.F, base - 1);
        ll t2 = query2(0, a[i].S.F);
        ll t3 = t1 + a[i].S.F * m;
        ll t4 = max(t2, t3) + a[i].S.S;
        ll t5 = query1(a[i].S.F, a[i].S.F) + a[i].S.F * m;
        ll t6 = query2(a[i].S.F, a[i].S.F);
        update1(a[i].S.F, max(t5, t4 - a[i].S.F * m));
        update2(a[i].S.F, max(t6, t4));
    }
    printf("%lld\n", max(query1(k, base - 1) + k * m, query2(0, k)));
    return 0;
}
Compilation message (stderr)
| # | Verdict | Execution time | Memory | Grader output | 
|---|---|---|---|---|
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