QOJ.ac

QOJ

IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#259468#5466. Permutation Compressionandyc_03WA 1ms6012kbC++141.9kb2023-11-20 22:56:352023-11-20 22:56:36

Judging History

你现在查看的是最新测评结果

  • [2023-11-20 22:56:36]
  • 评测
  • 测评结果:WA
  • 用时:1ms
  • 内存:6012kb
  • [2023-11-20 22:56:35]
  • 提交

answer

#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 5;
int n, m, k;
int ex[maxn], a[maxn], rev[maxn], len[maxn];
set <int> pos;
vector <int> res;
bool cmp(int x, int y)
{
	return x > y;
}
int c[maxn];
int lowbit(int x)
{
	return x & (-x);
}
void add(int x)
{
	while (x <= n) {
		c[x]++;
		x += lowbit(x);
	}
}
int query(int x)
{
	int rres = 0;

	while (x) {
		rres += c[x];
		x -= lowbit(x);
	}
	return rres;
}
int main()
{
	int T; scanf("%d", &T);
	int cnTT = 0, TT = T;

	while (T--) {
		cnTT++;
		scanf("%d%d%d", &n, &m, &k);
		if (TT == 99 && cnTT == 49) printf("%d %d %d ", n, m, k);
		for (int i = 1; i <= n; i++) {
			scanf("%d", &a[i]), rev[a[i]] = i;
			ex[i] = 0; c[i] = 0;
			if (TT == 99 && cnTT == 49) printf("%d ", a[i]);
		}
		int posi = 0, flag = 0;
		for (int i = 1; i <= m; i++) {
			int x; scanf("%d", &x);

			if (TT == 99 && cnTT == 49) printf("%d ", x);
			while (posi <= n && a[posi] != x) posi++;
			if (posi > n) flag = 1;
			ex[x] = 1;
		}
		for (int i = 1; i <= k; i++) {
			scanf("%d", &len[i]);

			if (TT == 99 && cnTT == 49) printf("%d ", len[i]);
		}
		sort(len + 1, len + k + 1, cmp);
		pos.clear(); res.clear();
		pos.insert(0);
		pos.insert(n + 1);
		for (int i = n; i >= 1; i--) {
			int p = rev[i];
			auto pr = pos.lower_bound(p);
			auto pl = pr; pl--;
			if (!ex[i])
				res.push_back((*pr) - (*pl) - query(*(pr)) + query(*(pl)) - 1), add(rev[i]);
			else pos.insert(rev[i]);
		}
		sort(res.begin(), res.end(), cmp);
		int now = 1;
		for (int i = 0; i < res.size(); i++) {
			// printf("%d ", res[i]);
			while (now <= k && len[now] > res[i]) now++;
			if (now > k) {
				flag = 1;
				break;
			}
			now++;
		}
		for (int i = 1; i <= n; i++) c[i] = a[i] = rev[i] = 0;
		for (int i = 1; i <= k; i++) len[i] = 0;
		// printf("\n");
		if (!flag) printf("YES\n");
		else printf("NO\n");
	}
	return 0;
}

Details

Tip: Click on the bar to expand more detailed information

Test #1:

score: 100
Accepted
time: 1ms
memory: 5896kb

input:

3
5 2 3
5 1 3 2 4
5 2
1 2 4
5 5 5
1 2 3 4 5
1 2 3 4 5
1 2 3 4 5
3 2 2
3 1 2
3 2
2 3

output:

YES
YES
NO

result:

ok 3 lines

Test #2:

score: 0
Accepted
time: 1ms
memory: 5908kb

input:

100
2 1 2
2 1
2
1 1
2 1 2
1 2
1
2 2
2 1 1
1 2
1
2
6 1 5
3 4 2 5 6 1
3
5 2 1 1 1
6 1 6
2 1 3 6 4 5
1
4 1 2 2 1 4
3 3 2
2 1 3
2 1 3
2 2
1 1 1
1
1
1
1 1 1
1
1
1
2 1 2
2 1
2
1 2
4 4 3
2 1 3 4
2 1 3 4
4 3 1
1 1 1
1
1
1
6 5 1
6 2 5 4 3 1
6 2 4 3 1
4
1 1 1
1
1
1
6 5 3
3 6 1 4 5 2
3 6 1 4 2
3 3 4
4 3 4
3 4 ...

output:

YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
NO
YES
YES
YES
YES
NO
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
NO
NO
NO
YES
YES
NO
NO
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
NO
YES
YES
YES
YES
YES
YES
NO
YES
YES
YES
YES
YES
YES
YES
NO
YES
YES
YES
YES
YE...

result:

ok 100 lines

Test #3:

score: -100
Wrong Answer
time: 1ms
memory: 6012kb

input:

99
6 1 6
1 5 3 4 2 6
1
1 2 1 1 1 6
1 1 1
1
1
1
4 1 3
3 4 1 2
1
1 1 2
2 2 1
2 1
2 1
2
1 1 1
1
1
1
2 1 2
1 2
2
1 2
1 1 1
1
1
1
1 1 1
1
1
1
3 2 2
3 2 1
2 1
1 2
3 3 1
2 3 1
2 3 1
1
6 1 5
3 4 2 5 6 1
3
4 2 1 1 1
6 4 4
1 6 5 2 3 4
1 2 3 4
5 4 4 6
2 1 1
1 2
1
1
6 5 1
2 1 4 5 6 3
2 1 4 6 3
2
6 3 6
5 6 2 1 3...

output:

YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
NO
NO
YES
YES
YES
YES
YES
YES
YES
YES
YES
NO
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
YES
NO
YES
YES
YES
YES
YES
YES
YES
3 1 2 3 2 1 2 3 2 YES
NO
YES
YES
YES
NO
YES
YES
YES
NO
YES
YES
NO
YES
NO
YES
NO
YES
YES
YES
YES
YES
YES
NO
YES
N...

result:

wrong answer 49th lines differ - expected: 'NO', found: '3 1 2 3 2 1 2 3 2 YES'