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답안 #309232

# 제출 시각 아이디 문제 언어 결과 실행 시간 메모리
309232 2020-10-02T22:53:07 Z VROOM_VARUN Skyscraper (JOI16_skyscraper) C++14
100 / 100
 68 ms 23924 KB 
// zscoder's not mine lmao
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef pair<int, int> ii;
typedef vector<int> vi;
typedef vector<ii> vii;
typedef long double ld;

#define fi first
#define se second
#define pb push_back
#define mp make_pair

ll dp[101][101][1001][3];
/*
dp[i][j][k][l] :
i - number of numbers placed
j - number of connected components
k - total sum currently (filling empty spaces with a_{i} (0-indexed)
l - number of endpoints that are filled
*/
ll a[101];
const ll MOD = 1e9 + 7;

int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, l;
  cin >> n >> l;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a, a + n);
  if (n == 1)  // special case
  {
    cout << 1;
    return 0;
  }
  a[n] = 10000;  // inf for simplicity
  if (a[1] - a[0] <= l)
    dp[1][1][a[1] - a[0]][1] =
        2;  // fill a[0] at one of the endpoints, there are 2 endpoints to fill.
  if (2 * (a[1] - a[0]) <= l)
    dp[1][1][2 * (a[1] - a[0])][0] =
        1;  // fill a[0] in the middle, positions doesn't matter.

  for (int i = 1; i < n; i++) {
    int diff = a[i + 1] - a[i];  // this thing is "INF" if i = n - 1.
    for (int j = 1; j <= i; j++) {
      for (int k = 0; k <= l; k++) {
        for (int z = 0; z < 3; z++) {
          if (!dp[i][j][k][z]) continue;  // this value does not exist
          // First, we try to fill one of the ends
          if (z < 2 &&
              k + diff * (2 * j - z - 1) <=
                  l)  // there are 2*j - z - 1 positions that we're supposed to
                      // "upgrade" (-1 because one of the positions is merged
                      // with the endpoints after this move)
          {
            if (i == n - 1) {
              dp[i + 1][j][k + diff * (2 * j - z - 1)][z + 1] =
                  (dp[i + 1][j][k + diff * (2 * j - z - 1)][z + 1] +
                   dp[i][j][k][z] * (1) * j) %
                  MOD;  // we have j con. comp. to choose to merge with
            } else if (z == 0 ||
                       j > 1)  // otherwise this coincides with i == n - 1
            {
              dp[i + 1][j][k + diff * (2 * j - z - 1)][z + 1] =
                  (dp[i + 1][j][k + diff * (2 * j - z - 1)][z + 1] +
                   dp[i][j][k][z] * (2 - z) * (j - z)) %
                  MOD;  // can only merge with the con comp. that are not
                        // connected to ends.
            }
            if (k + diff * (2 * j - z + 1) <= l)  // now we create a new cc.
            {
              dp[i + 1][j + 1][k + diff * (2 * j - z + 1)][z + 1] =
                  (dp[i + 1][j + 1][k + diff * (2 * j - z + 1)][z + 1] +
                   dp[i][j][k][z] * (2 - z)) %
                  MOD;  // we can choose one of the ends to create
            }
          }
          // Next, we dont fill the ends.
          // Part 1 : Create new cc
          if (k + diff * (2 * j - z + 2) <= l)  // 2 new positions to "upgrade"
          {
            dp[i + 1][j + 1][k + diff * (2 * j - z + 2)][z] =
                (dp[i + 1][j + 1][k + diff * (2 * j - z + 2)][z] +
                 dp[i][j][k][z]) %
                MOD;  // nothing new happens
          }
          // Part 2 : Stick to one cc
          if (k + diff * (2 * j - z) <= l)  // no new positions to "upgrade"
          {
            dp[i + 1][j][k + diff * (2 * j - z)][z] =
                (dp[i + 1][j][k + diff * (2 * j - z)][z] +
                 dp[i][j][k][z] * (2 * j - z)) %
                MOD;  // we can merge in 2*j - z possible positions
          }
          // Part 3 : Merge two ccs together
          if ((k + diff * (2 * j - z - 2) <= l) && (j >= 2) &&
              (i == n - 1 || j > 2 || z < 2)) {
            if (z == 0) {
              dp[i + 1][j - 1][k + diff * (2 * j - z - 2)][z] =
                  (dp[i + 1][j - 1][k + diff * (2 * j - z - 2)][z] +
                   dp[i][j][k][z] * j * (j - 1)) %
                  MOD;  // there are jP2 possible merges
            }
            if (z == 1) {
              dp[i + 1][j - 1][k + diff * (2 * j - z - 2)][z] =
                  (dp[i + 1][j - 1][k + diff * (2 * j - z - 2)][z] +
                   dp[i][j][k][z] * (j - 1) * (j - 1)) %
                  MOD;  // there are (j-1)P2+(j-1) merges
            }
            if (z == 2) {
              if (i == n - 1) {
                dp[i + 1][j - 1][k + diff * (2 * j - z - 2)][z] =
                    (dp[i + 1][j - 1][k + diff * (2 * j - z - 2)][z] +
                     dp[i][j][k][z]) %
                    MOD;  // there's only 1 place it can go.
              } else {
                dp[i + 1][j - 1][k + diff * (2 * j - z - 2)][z] =
                    (dp[i + 1][j - 1][k + diff * (2 * j - z - 2)][z] +
                     dp[i][j][k][z] * (j - 2) * (j - 1)) %
                    MOD;  // there're (j-2)P2 + 2(j-2) possiblilities
              }
            }
          }
        }
      }
    }
  }

  ll answer = 0;
  for (int i = 0; i <= l; i++) {
    answer = (answer + dp[n][1][i][2]) %
             MOD;  // sum the dp values for all possible sums
  }
  cout << answer << '\n';
  return 0;
}

Subtask #1
5.0 / 5.0

# 결과 실행 시간 메모리 Grader output
1 Correct 1 ms 384 KB Output is correct
2 Correct 0 ms 384 KB Output is correct
3 Correct 1 ms 384 KB Output is correct
4 Correct 1 ms 384 KB Output is correct
5 Correct 1 ms 512 KB Output is correct
6 Correct 1 ms 512 KB Output is correct
7 Correct 1 ms 384 KB Output is correct
8 Correct 1 ms 512 KB Output is correct
9 Correct 1 ms 512 KB Output is correct
10 Correct 1 ms 512 KB Output is correct

Subtask #2
15.0 / 15.0

# 결과 실행 시간 메모리 Grader output
1 Correct 1 ms 640 KB Output is correct
2 Correct 1 ms 640 KB Output is correct
3 Correct 1 ms 640 KB Output is correct
4 Correct 1 ms 512 KB Output is correct
5 Correct 1 ms 512 KB Output is correct
6 Correct 1 ms 640 KB Output is correct
7 Correct 1 ms 640 KB Output is correct
8 Correct 1 ms 640 KB Output is correct
9 Correct 1 ms 768 KB Output is correct
10 Correct 1 ms 640 KB Output is correct

Subtask #3
80.0 / 80.0

# 결과 실행 시간 메모리 Grader output
1 Correct 1 ms 384 KB Output is correct
2 Correct 0 ms 384 KB Output is correct
3 Correct 1 ms 384 KB Output is correct
4 Correct 1 ms 384 KB Output is correct
5 Correct 1 ms 512 KB Output is correct
6 Correct 1 ms 512 KB Output is correct
7 Correct 1 ms 384 KB Output is correct
8 Correct 1 ms 512 KB Output is correct
9 Correct 1 ms 512 KB Output is correct
10 Correct 1 ms 512 KB Output is correct
11 Correct 1 ms 640 KB Output is correct
12 Correct 1 ms 640 KB Output is correct
13 Correct 1 ms 640 KB Output is correct
14 Correct 1 ms 512 KB Output is correct
15 Correct 1 ms 512 KB Output is correct
16 Correct 1 ms 640 KB Output is correct
17 Correct 1 ms 640 KB Output is correct
18 Correct 1 ms 640 KB Output is correct
19 Correct 1 ms 768 KB Output is correct
20 Correct 1 ms 640 KB Output is correct
21 Correct 3 ms 1280 KB Output is correct
22 Correct 68 ms 23924 KB Output is correct
23 Correct 56 ms 8184 KB Output is correct
24 Correct 55 ms 12152 KB Output is correct
25 Correct 59 ms 9336 KB Output is correct
26 Correct 54 ms 8568 KB Output is correct
27 Correct 25 ms 9728 KB Output is correct
28 Correct 32 ms 11904 KB Output is correct
29 Correct 56 ms 16504 KB Output is correct
30 Correct 58 ms 9464 KB Output is correct