QOJ.ac

QOJ

IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#745576#8780. Training, Round 2cyj888WA 8ms127248kbC++142.0kb2024-11-14 10:38:492024-11-14 10:38:53

Judging History

你现在查看的是最新测评结果

  • [2024-11-14 10:38:53]
  • 评测
  • 测评结果:WA
  • 用时:8ms
  • 内存:127248kb
  • [2024-11-14 10:38:49]
  • 提交

answer

#include <bits/stdc++.h>
//#define int long long
#define fi first
#define se second 
#define pb push_back
#define ott(i, l, r) for (register int i = (l); i <= (r); i ++)
#define tto(i, l, r) for (register int i = (r); i >= (l); i --)
using namespace std;
typedef long long ll;
typedef long double ld;
int read () {
	int x = 0; bool f = false; char c = getchar ();
	while (!isdigit (c)) f |= (c == '-'), c = getchar ();
	while (isdigit (c)) x = (x << 3) + (x << 1) + (c ^ 48), c = getchar ();
	return f ? -x : x;
}
const int N = 5500, mod = 1e9 + 7;
int n, s1, s2, res;
int vis[N][N]; bool ins[N][N];
vector <pair <int, int> > cac;
inline void add (const int &x, const int &y) {
	if (!vis[x + 1][y]) cac.pb ({x + 1, y}), vis[x + 1][y] = true;
	if (!vis[x][y + 1]) cac.pb ({x, y + 1}), vis[x][y + 1] = true;
	return ;
}
struct Segment_tree {
	int cnt, id; int siz[N];
	#define ls p << 1
	#define rs p << 1 | 1
	void pup (int p) {
		return void (siz[p] = siz[ls] + siz[rs]);
	}
	void ins (int x, int p = 1, int l = 0, int r = n) {
		if (l >= r) return void (++ siz[p]);
		int mid = l + r >> 1;
		x <= mid ? ins (x, ls, l, mid) : ins (x, rs, mid + 1, r);
		return pup (p);
	}
	void del (int x, int y, int p = 1, int l = 0, int r = n) {
		if (!siz[p]) return ;
		if (l >= r) {
			add (id, l); return void (-- siz[p]);
		}
		int mid = l + r >> 1;
		if (x <= mid) del (x, y, ls, l, mid); if (y > mid) del (x, y, rs, mid + 1, r);
		return pup (p);
	}
} has[N];
int main () {
	n = read (), s1 = read (), s2 = read ();
	ott (i, 0, n) has[i].id = i;
	vis[0][0] = true, has[0].ins (0);
	ott (i, 1, n) {
		register int l1 = read () - s1, r1 = read () - s1, l2 = read () - s2, r2 = read () - s2;
		if (l1 < 0) l1 = 0; if (l2 < 0) l2 = 0; if (r1 > n) r1 = n; if (r2 > n) r2 = n;
		if (l2 <= r2) ott (x, l1, r1) has[x].del (l2, r2);
		for (auto &[x, y] : cac) has[x].ins (y); cac.clear ();
	}
	ott (i, 0, n)
		ott (j, 0, n - i)
		    if (vis[i][j]) 
		    	res = max (res, i + j);
	printf ("%d\n", res);
    return 0;
}

Details

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Test #1:

score: 100
Accepted
time: 0ms
memory: 7944kb

input:

3 0 0
0 1 0 1
1 1 0 1
1 1 1 1

output:

3

result:

ok single line: '3'

Test #2:

score: 0
Accepted
time: 7ms
memory: 127248kb

input:

5000 801577551 932138594
801577551 801577551 932138594 932138594
801577552 801577552 932138594 932138594
801577552 801577552 932138595 932138595
801577552 801577552 932138596 932138596
801577553 801577553 932138596 932138596
801577553 801577553 932138597 932138597
801577553 801577553 932138598 93213...

output:

5000

result:

ok single line: '5000'

Test #3:

score: -100
Wrong Answer
time: 8ms
memory: 122444kb

input:

5000 932138594 801577551
932138594 932138594 801577551 801577551
932138594 932138594 801577552 801577552
932138595 932138595 801577552 801577552
932138596 932138596 801577552 801577552
932138596 932138596 801577553 801577553
932138597 932138597 801577553 801577553
932138598 932138598 801577553 80157...

output:

3267

result:

wrong answer 1st lines differ - expected: '5000', found: '3267'