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IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#216631#5414. Stop, Yesterday Please No MoreddWA 6ms8144kbC++202.1kb2023-10-15 20:40:022023-10-15 20:40:03

Judging History

你现在查看的是最新测评结果

  • [2023-10-15 20:40:03]
  • 评测
  • 测评结果:WA
  • 用时:6ms
  • 内存:8144kb
  • [2023-10-15 20:40:02]
  • 提交

answer

#include <bits/stdc++.h>

using namespace std;

const int N = 2e6 + 10, M = 2e3 + 10;

int n, m, k;
set<pair<int, int> > a;
char st[N];
int sum[M][M];
int main() {
	int T; scanf("%d", &T);
	while (T--) {
		a.clear();
		scanf("%d%d%d", &n, &m, &k);
		for (int i = 0; i <= n; ++i)
			for (int j = 0; j <= m; ++j)
				sum[i][j] = 0;
		scanf("%s", st + 1);
		int U = 0, D = 0, L = 0, R = 0, x = 0, y = 0;
		int len = strlen(st + 1);
		for (int i = 1; i <= len; ++i) {
			if (st[i] == 'D')
				++x;
			if (st[i] == 'U')
				--x;
			if (st[i] == 'L')
				--y;
			if (st[i] == 'R')
				++y;
			a.insert(make_pair(-x, -y));
			U = max(U, -x);
			D = max(D, x);
			L = max(L, -y);
			R = max(R, y);
		}
	//	cout << U << ' ' << D <<' ' <<L << ' ' << R << endl;
		if (U + D >= n || L + R >= m) {
			if (k == 0)
				printf("%d\n", n * m);
			else
				printf("0\n");
			continue;
		}
	//	cout << U << ' ' << D << ' ' << L << ' ' << R << endl;
		++sum[U + 1][L + 1];
		--sum[n - D + 1][L + 1];
		--sum[U + 1][m - R + 1];
		++sum[n - D + 1][m - R + 1];
		for (set<pair<int,int> >::iterator it = a.begin(); it != a.end(); it++) {
			int xx1 = U + 1 - it->first, yy1 = L + 1 - it->second;
			int xx2 = n - D - it->first, yy2 = m - R - it->second;
			xx1 = max(xx1, 1);
			yy1 = max(yy1, 1);
			xx2 = min(xx2, n);
			yy2 = min(yy2, m);
			if (xx1 > n || yy1 > m || xx2 < 1 || yy2 < 1)
				continue;
		//	cout << it->first <<' ' <<it->second << endl;
		//	cout << xx1 << ' ' << yy1 <<' ' << xx2 << ' ' << yy2 << endl;
			++sum[xx1][yy1];
			--sum[xx2 + 1][yy1];
			--sum[xx1][yy2 + 1];
			++sum[xx2 + 1][yy2 + 1];
		}
		int ans = 0;
		for (int i = 1; i <= n; ++i)
			for (int j = 1; j <= m; ++j) {
				sum[i][j] += sum[i - 1][j] + sum[i][j - 1] - sum[i - 1][j - 1];
				if ((n - U - D) * (m - L - R) - sum[i][j] == k)
					++ans;
			}
	/*	for (int i = 1; i <= n; ++i) {
			for (int j = 1; j <= m; ++j)
				cout << (n - U - D) * (m - L - R) - sum[i][j] << ' ';
			cout << endl;
		}*/
		printf("%d\n", ans);
	}
	return 0;
}
/*
2
4 5 15
U
4 5 14
U
*/

Details

Tip: Click on the bar to expand more detailed information

Test #1:

score: 100
Accepted
time: 1ms
memory: 5848kb

input:

3
4 5 3
ULDDRR
4 5 0
UUUUUUU
4 5 10
UUUUUUU

output:

2
20
0

result:

ok 3 number(s): "2 20 0"

Test #2:

score: -100
Wrong Answer
time: 6ms
memory: 8144kb

input:

1060
19 12 0
UDLDDUUUUDDDLLRDUDUURULUUUDRDUDRDRLRLRLULULLLDLDDRLUUUURUUUDDRLLRUUUDULURUULLRDRLRDDURDUUURRRLURLRUULRRUDURDLUUURDLURDDLUUURDDRLLURRDLRUDLRDRLLRRDRDDLDRURRRLUDULLLRUUDLRRURRDLLRRRDLLRDDDLRLRURURDDDL
11 1 0
UR
3 18 33
UDRLR
17 11 132
RLDRDLDRUU
6 10 13
UULUDDLRDLUUDLDD
1 15 0
D
6 20 50
D...

output:

228
11
19
0
18
15
34
240
15
0
0
13
14
18
26
16
1
19
108
8
1
2
3
7
0
29
14
21
0
8
1
9
13
5
320
3
7
3
0
0
12
0
11
0
0
14
0
22
36
51
23
5
6
3
11
48
28
8
7
22
49
13
10
2
0
4
18
44
0
14
9
0
4
30
14
99
105
5
14
17
0
66
10
11
28
52
34
8
33
14
0
90
0
0
11
3
48
30
36
12
0
30
7
8
3
11
16
45
20
34
0
20
0
21
0
...

result:

wrong answer 3rd numbers differ - expected: '20', found: '19'