QOJ.ac

QOJ

ID题目提交者结果用时内存语言文件大小提交时间测评时间
#323097#7733. Cool, It’s Yesterday Four Times MorezlxFTH#WA 4ms7804kbC++141.9kb2024-02-08 14:34:462024-02-08 14:34:46

Judging History

你现在查看的是最新测评结果

  • [2024-02-08 14:34:46]
  • 评测
  • 测评结果:WA
  • 用时:4ms
  • 内存:7804kb
  • [2024-02-08 14:34:46]
  • 提交

answer

// QOJ7733 Cool, It’s Yesterday Four Times More
#include <cstdio>

const int N = 5e3 + 5;
const int dx[] = {0, 0, 1, -1};
const int dy[] = {1, -1, 0, 0};

int n, m, tot;
int f[N][N];
char s[N][N];
struct Node {
  int i, j, x, y;
} q[N];

inline int id(int x, int y) {
  return x * (m + 2) + y;
}

void solve() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) scanf("%s", s[i] + 1);
  for (int i = 0; i <= n + 1; ++i)
    for (int j = 0; j <= m + 1; ++j) {
      if (!i || !j || i > n || j > m) s[i][j] = 'O';
    }
  tot = (n + 2) * (m + 2);
  for (int i = 0; i < tot; ++i)
    for (int j = 0; j < tot; ++j) f[i][j] = 0;
  int h = 1, t = 0;
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= m; ++j) if (s[i][j] == '.') {
      for (int x = 0; x <= n + 1; ++x) {
        for (int y = 0; y <= m + 1; ++y) if (s[x][y] == 'O') {
          f[id(i, j)][id(x, y)] = 1;
          q[++t] = {i, j, x, y};
        }
      }
    }
  while (h <= t) {
    int i = q[h].i, j = q[h].j, x = q[h].x, y = q[h].y;
    h++;
    for (int k = 0; k < 4; ++k) {
      int ni = i + dx[k];
      int nj = j + dy[k];
      int nx = x + dx[k];
      int ny = y + dy[k];
      if (s[ni][nj] == '.' && nx >= 0 && nx <= n + 1
          && ny >= 0 && ny <= m + 1 && !f[id(ni, nj)][id(nx, ny)]) {
        f[id(ni, nj)][id(nx, ny)] = 1; 
        q[++t] = {ni, nj, nx, ny};
      }
    }
  }
  int ans = 0;
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= m; ++j) if (s[i][j] == '.') {
      int fl = 1;
      for (int x = 0; x <= n + 1; ++x) {
        for (int y = 0; y <= m + 1; ++y) if (s[x][y] == '.') {
          if (x == i && y == j) continue;
          fl &= f[id(i, j)][id(x, y)];
          if (!fl) break;
        }
        if (!fl) break;
      }
      ans += fl;
    }
  printf("%d\n", ans);
}

int main() {
  int t;
  scanf("%d", &t);
  while (t--) solve();
  return 0;
}

详细

Test #1:

score: 100
Accepted
time: 0ms
memory: 3604kb

input:

4
2 5
.OO..
O..O.
1 3
O.O
1 3
.O.
2 3
OOO
OOO

output:

3
1
0
0

result:

ok 4 lines

Test #2:

score: 0
Accepted
time: 1ms
memory: 5676kb

input:

200
2 4
OOO.
OO..
2 3
OOO
.O.
3 3
O.O
OOO
OO.
4 1
.
.
O
O
1 2
.O
1 1
.
2 5
.OO..
.O.O.
2 1
O
O
1 1
O
1 3
.OO
5 1
O
O
.
O
.
5 2
O.
..
O.
.O
..
5 3
...
...
.OO
..O
OOO
3 5
..O.O
.O.O.
.OO.O
5 2
.O
OO
O.
O.
..
2 1
O
O
3 5
.O.OO
O...O
..OO.
1 5
.....
5 1
O
.
O
.
.
5 3
OOO
OO.
.OO
OO.
O.O
2 1
O
.
5 2
O.
...

output:

3
0
0
2
1
1
3
0
0
1
0
7
9
4
4
0
6
5
2
0
1
6
4
5
2
0
0
5
3
3
1
4
1
0
7
5
2
3
7
3
0
6
2
2
2
0
4
6
6
3
3
2
3
5
2
1
0
3
3
4
4
2
2
0
7
6
4
8
5
3
2
5
2
1
2
1
4
0
0
2
5
1
4
6
6
1
6
2
2
3
4
5
2
1
0
1
9
3
4
11
0
3
2
1
0
0
4
3
1
4
3
8
3
0
3
6
2
5
1
3
3
4
0
2
11
2
2
4
0
4
4
6
2
1
2
3
0
5
0
16
4
3
2
6
0
8
3
3
1...

result:

ok 200 lines

Test #3:

score: -100
Wrong Answer
time: 4ms
memory: 7804kb

input:

50
10 9
OOOO.O...
O...O.OOO
.....O...
..OO..O.O
...O..O.O
..OOO..O.
..OOO...O
.OO.O..OO
.O.O.OO..
.O..O.O.O
10 10
.O.OOO.OO.
...OOOOOOO
...O.O..O.
.O.O..O...
.O.O.OO..O
..OO.O..OO
O....O..OO
OO...OOOOO
OO.OO.O..O
.O.O.OOOOO
10 8
O..OOO.O
O.OOOO..
O..OO.OO
OO..OO..
.OOO..O.
.OO.OO.O
OOO..OO.
..O..OO....

output:

23
21
13
25
26
37
27
29
20
26
25
29
21
29
21
11
32
31
33
34
25
31
18
25
41
25
20
31
20
29
18
21
27
28
12
13
20
17
32
29
28
23
23
23
24
18
28
17
21
24

result:

wrong answer 1st lines differ - expected: '31', found: '23'