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IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#225502#7523. Partially Free MealPetroTarnavskyi#WA 234ms25800kbC++172.7kb2023-10-24 18:44:522023-10-24 18:44:52

Judging History

你现在查看的是最新测评结果

  • [2023-10-24 18:44:52]
  • 评测
  • 测评结果:WA
  • 用时:234ms
  • 内存:25800kb
  • [2023-10-24 18:44:52]
  • 提交

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 (a[idxes[i]] + b[idxes[i]] < mn)
		{
			mn = a[idxes[i]] + 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(SZ(sSmall1) == k && SZ(sSmall2) == k);
			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)
	{
		assert(SZ(sSmall1) == k);
		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;
}

Details

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Test #1:

score: 100
Accepted
time: 0ms
memory: 3884kb

input:

3
2 5
4 3
3 7

output:

7
11
16

result:

ok 3 lines

Test #2:

score: -100
Wrong Answer
time: 234ms
memory: 25800kb

input:

200000
466436993 804989151
660995237 756645598
432103296 703610564
6889895 53276988
873617076 822481192
532911431 126844295
623111499 456772252
937464699 762157133
708503076 786039753
78556972 5436013
582960979 398984169
786333369 325119902
930705057 615928139
924915828 506145001
164984329 208212435...

output:

1318594
3208018
5570526
7340845
9223347
11487338
12898863
14697855
14788280
16172895
17768627
19336633
20693779
22005236
23389851
25073157
26760338
28509336
30294967
32093959
33976461
35893858
37754030
39588384
41470886
43388283
45309771
47573762
49838235
52476909
55158583
58072674
61026526
64003437...

result:

wrong answer 6th lines differ - expected: '11149865', found: '11487338'