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QOJ
ID | 题目 | 提交者 | 结果 | 用时 | 内存 | 语言 | 文件大小 | 提交时间 | 测评时间 |
---|---|---|---|---|---|---|---|---|---|
#225496 | #7521. Find the Gap | PetroTarnavskyi# | WA | 0ms | 3552kb | C++17 | 2.5kb | 2023-10-24 18:33:08 | 2023-10-24 18:33:09 |
Judging History
answer
#include <bits/stdc++.h>
using namespace std;
#define FOR(i, a, b) for(int i = (a); i < (b); i++)
#define RFOR(i, a, b) for(int i = (a) - 1; i >= (b); i--)
#define SZ(a) (int)a.size()
#define ALL(a) a.begin(), a.end()
#define PB push_back
#define MP make_pair
#define F first
#define S second
typedef long long LL;
typedef vector<int> VI;
typedef pair<int, int> PII;
typedef double db;
const LL LINF = 1e18;
int main()
{
ios::sync_with_stdio(0);
cin.tie(0);
cout << fixed << setprecision(15);
int n;
cin >> n;
VI a(n), b(n), idxes(n);
FOR(i, 0, n)
cin >> a[i] >> b[i];
iota(ALL(idxes), 0);
sort(ALL(idxes), [&](int i, int j) {return b[i] < b[j];});
vector<int> vec;
int mn = INT_MAX;
RFOR(i, n, 0)
{
if (b[idxes[i]] < mn)
{
mn = b[idxes[i]];
vec.PB(i);
}
}
reverse(ALL(vec));
vector<LL> ans(n);
int k = 0;
multiset<int> sBig1, sSmall1, sBig2, sSmall2;
LL sum1 = a[idxes[vec[0]]] + b[idxes[vec[0]]], sum2 = 0;
FOR(j, 0, vec[0])
sBig1.insert(a[idxes[j]]);
if (SZ(vec) > 1)
{
FOR(j, 0, vec[1])
sBig2.insert(a[idxes[j]]);
sum2 = a[idxes[vec[1]]] + b[idxes[vec[1]]];
}
ans[0] = sum1;
FOR(i, 0, SZ(vec) - 1)
{
while (k + 1 < n)
{
assert(!sBig2.empty());
LL nsum1 = sum1 + (sBig1.empty() ? LINF : *sBig1.begin());
LL nsum2 = sum2 + *sBig2.begin();
if (nsum2 < nsum1)
break;
assert(!sBig1.empty());
sum1 += *sBig1.begin();
sSmall1.insert(*sBig1.begin());
sBig1.erase(sBig1.begin());
sum2 += *sBig2.begin();
sSmall2.insert(*sBig2.begin());
sBig2.erase(sBig2.begin());
k++;
ans[k] = sum1;
}
sum1 += a[idxes[vec[i + 1]]] + b[idxes[vec[i + 1]]] - (a[idxes[vec[i]]] + b[idxes[vec[i]]]);
FOR(j, vec[i], vec[i + 1])
{
sum1 += a[idxes[j]];
sSmall1.insert(a[idxes[j]]);
}
while (SZ(sSmall1) > k)
{
sum1 -= *prev(sSmall1.end());
sBig1.insert(*prev(sSmall1.end()));
sSmall1.erase(prev(sSmall1.end()));
}
if (i + 2 == SZ(vec))
break;
sum2 += a[idxes[vec[i + 2]]] + b[idxes[vec[i + 2]]] - (a[idxes[vec[i + 1]]] + b[idxes[vec[i + 1]]]);
FOR(j, vec[i + 1], vec[i + 2])
{
sum2 += a[idxes[j]];
sSmall2.insert(a[idxes[j]]);
}
while (SZ(sSmall2) > k)
{
sum2 -= *prev(sSmall2.end());
sBig2.insert(*prev(sSmall2.end()));
sSmall2.erase(prev(sSmall2.end()));
}
}
while (k + 1 < n)
{
sum1 += *sBig1.begin();
sSmall1.insert(*sBig1.begin());
sBig1.erase(sBig1.begin());
k++;
ans[k] = sum1;
}
FOR(i, 0, n)
cout << ans[i] << "\n";
return 0;
}
詳細信息
Test #1:
score: 0
Wrong Answer
time: 0ms
memory: 3552kb
input:
8 1 1 1 1 1 2 1 2 1 1 2 2 2 1 1 2 1 2 2 2 1 2 2 2
output:
3 4 5 6 7 8 10 12
result:
wrong answer 1st numbers differ - expected: '1.0000000', found: '3.0000000', error = '2.0000000'