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IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#216439#6439. Cloud Retainer's Gameveg#AC ✓1487ms156364kbC++142.0kb2023-10-15 18:18:492023-10-15 18:18:49

Judging History

你现在查看的是最新测评结果

  • [2023-10-15 18:18:49]
  • 评测
  • 测评结果:AC
  • 用时:1487ms
  • 内存:156364kb
  • [2023-10-15 18:18:49]
  • 提交

answer

#include <bits/stdc++.h>
using namespace std;
struct AAA {
	int x, y, t;
} a[400005];
int dp[400005][2];
map<int, int> M1, M2;
int ncnt, H;
struct Node {
	int ls, rs, mx;
} t[13000005];
void add(int &p, int l, int r, int x, int y) {
	if (!p) {
		p = ++ncnt;
		t[p].ls = t[p].rs = 0;
		t[p].mx = -1e9;
	}
	t[p].mx = max(t[p].mx, y);
	if (l == r) return;
	int mid = l + (r - l) / 2;
	if (x <= mid) add(t[p].ls, l, mid, x, y);
	else add(t[p].rs, mid + 1, r, x, y);
}
int ask(int p, int l, int r, int x) {
	if (!p || l > x) return -1e9;
	if (r <= x) return t[p].mx;
	int mid = l + (r - l) / 2;
	return max(ask(t[p].ls, l, mid, x), ask(t[p].rs, mid + 1, r, x));
}
void ins(int i) {
//	printf("M1:mod<%d> %d %d\n", (a[i].x - a[i].y + H) % H, a[i].x - a[i].y, dp[i][0]);
//	printf("M2:mod<%d> %d %d\n", (a[i].x + a[i].y) % H, a[i].x + a[i].y, dp[i][1]);
	add(M1[(a[i].x - a[i].y + H) % H], -1e9, 1e9, a[i].x - a[i].y, dp[i][0]);
	add(M2[(a[i].x + a[i].y) % H], 0, 2e9, a[i].x + a[i].y, dp[i][1]);
}
int main() {
	int T;
	cin >> T;
	while (T--) {
		M1.clear();
		M2.clear();
		ncnt = 0;
		int n, m;
		scanf("%d", &H);
		H+=H;
		scanf("%d", &n);
		for (int i = 1; i <= n; ++i) {
			int x, y;
			scanf("%d%d", &x, &y);
			a[i] = (AAA){x, y, 0};
		}
		scanf("%d", &m);
		for (int i = 1; i <= m; ++i) {
			int x, y;
			scanf("%d%d", &x, &y);
			a[i + n] = (AAA){x, y, 1};
		}
		sort(a + 1, a + n + m + 1, [](const AAA &a, const AAA &b) { return a.x < b.x; });
		ins(0);
		int ans = 0;
		for (int i = 1; i <= n + m; ++i) {
			dp[i][1] = a[i].t + max(ask(M1[(a[i].x + a[i].y) % H], -1e9, 1e9, a[i].x + a[i].y),
							ask(M2[(a[i].x + a[i].y) % H], 0, 2e9, a[i].x + a[i].y));
			dp[i][0] = a[i].t + max(ask(M1[(a[i].x - a[i].y + H) % H], -1e9, 1e9, a[i].x - a[i].y),
							ask(M2[(a[i].x - a[i].y + H) % H], 0, 2e9, a[i].x - a[i].y));
			ans = max(ans, dp[i][1]);
			ans = max(ans, dp[i][0]);
			if (!a[i].t) dp[i][0] = dp[i][1] = max(dp[i][0], dp[i][1]);
			ins(i);
		}
		printf("%d\n", ans);
	}
	return 0;
}

Details

Tip: Click on the bar to expand more detailed information

Test #1:

score: 100
Accepted
time: 1ms
memory: 7980kb

input:

2
4
3
1 1
2 2
6 2
4
3 1
3 3
5 1
7 3
3
1
4 2
3
1 1
6 2
9 1

output:

3
3

result:

ok 2 number(s): "3 3"

Test #2:

score: 0
Accepted
time: 1117ms
memory: 17120kb

input:

5503
10
19
2 4
2 8
8 3
8 4
8 7
2 7
2 6
1 5
3 2
6 4
2 1
4 5
2 5
7 1
4 7
5 7
2 2
8 6
8 1
12
5 1
4 8
5 2
6 1
3 6
1 1
1 7
7 2
5 6
6 8
1 2
3 5
10
5
9 5
10 7
6 6
5 7
1 3
9
6 8
8 8
6 4
2 9
5 4
4 2
10 9
2 3
2 1
7
1
4 3
14
4 6
6 1
2 1
7 6
2 3
4 4
5 3
6 5
1 4
3 4
3 2
6 2
8 6
8 2
6
6
5 2
5 1
3 1
2 3
7 4
5 5
3
...

output:

2
1
2
1
3
2
0
2
4
6
1
2
0
0
1
2
1
1
0
1
0
0
2
1
1
3
2
3
3
2
1
2
0
1
5
1
1
1
0
1
3
1
2
3
3
3
2
1
0
3
1
2
2
0
4
1
1
0
1
2
2
2
1
1
1
1
2
3
2
2
2
1
1
3
1
3
0
0
3
4
5
1
1
1
1
1
0
2
0
0
3
0
2
1
1
1
0
3
2
1
3
4
3
2
2
4
2
4
2
1
2
1
0
1
3
0
3
0
2
1
0
2
5
1
2
2
1
0
1
3
0
2
3
1
4
2
2
0
2
3
2
0
0
3
1
1
1
1
3
2
...

result:

ok 5503 numbers

Test #3:

score: 0
Accepted
time: 1487ms
memory: 156364kb

input:

54
83
1995
54 14
42 63
23 55
46 52
94 71
16 18
51 54
62 47
90 38
42 50
82 20
8 28
52 64
49 19
56 5
10 74
99 30
90 42
48 2
11 78
4 38
78 77
26 26
47 12
82 60
41 17
87 2
37 16
51 15
32 63
88 82
76 33
44 10
94 28
31 5
30 80
29 19
35 70
88 78
39 69
40 5
84 52
87 59
54 36
34 76
88 42
42 37
79 70
27 77
19...

output:

47
32
32
32
38
32
39
33
39
40
36
32
36
32
46
30
35
41
40
36
108
90
98
81
166
115
106
170
148
113
198
72
57
202
337
153
186
978
87
886
151
489
111
112
90
154
174
188
266
59
10210
1041
87
981

result:

ok 54 numbers