Submission #332034

#	Time	Username	Problem	Language	Result	Execution time	Memory
332034		dolphingarlic	Skyscraper (JOI16_skyscraper)	C++14	0 / 100	1 ms	748 KiB

skyscraper

#include <bits/stdc++.h>
typedef long long ll;
using namespace std;

const ll MOD = 1e9 + 7;

int a[102];
ll dp[101][101][1001][3];
// dp[i][j][k][m] = No. of ways to insert first i buildings into permutation with
//                  j connected components, total cost k, and m ends of permutation inserted

int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int n, l;
    cin >> n >> l;
    for (int i = 1; i <= n; i++) cin >> a[i];
    sort(a + 1, a + n + 1);
    a[n + 1] = 10000;
    dp[0][0][0][0] = 1;
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= i; j++) {
            for (int k = 0; k <= l; k++) {
                for (int m = 0; m <= 2; m++) {
                    int cost_diff = (2 * j - m) * (a[i + 1] - a[i]);
                    if (cost_diff > k || i + j + 1 - m > n) continue;
                    // Case 1: we form a new connected component in the middle
                    dp[i][j][k][m] += dp[i - 1][j - 1][k - cost_diff][m];
                    dp[i][j][k][m] %= MOD;
                    // Case 2: we form a new connected component at the end
                    if (m) dp[i][j][k][m] += (3 - m) * dp[i - 1][j - 1][k - cost_diff][m - 1];
                    dp[i][j][k][m] %= MOD;
                    // Case 3: we append a[i] to the end of a component so that it isn't an end of the permutation
                    dp[i][j][k][m] += (2 * j - m) * dp[i - 1][j][k - cost_diff][m];
                    dp[i][j][k][m] %= MOD;
                    // Case 4: we append a[i] to the end of a component so that it's an end of the permutation
                    if (m == 1) dp[i][j][k][m] += 2 * j * dp[i - 1][j][k - cost_diff][m - 1];
                    if (m == 2) {
                        if (i == n) dp[i][j][k][m] += dp[i - 1][j][k - cost_diff][m - 1];
                        else if (j > 1) dp[i][j][k][m] += (j - 1) * dp[i - 1][j][k - cost_diff][m - 1];
                    }
                    dp[i][j][k][m] %= MOD;
                    // Case 5: we join two components
                    if (m == 2) {
                        if (i == n) 
                            dp[i][j][k][m] += dp[i - 1][j + 1][k - cost_diff][m];
                        else
                            dp[i][j][k][m] += j * (j - 1) * dp[i - 1][j + 1][k - cost_diff][m];
                    } else if (m == 1)
                        dp[i][j][k][m] += j * j * dp[i - 1][j + 1][k - cost_diff][m]; 
                    else
                        dp[i][j][k][m] += j * (j + 1) * dp[i - 1][j + 1][k - cost_diff][m];
                    dp[i][j][k][m] %= MOD;
                    // if (dp[i][j][k][m]) cout << i << ' ' << j << ' ' << k << ' ' << m << ' ' << dp[i][j][k][m] << '\n';
                }
            }
        }
    }

    ll ans = 0;
    for (int i = 0; i <= l; i++) ans += dp[n][1][i][2];
    cout << ans;
    return 0;
}

• Subtask 1: 0 / 5
• Subtask 2: 0 / 15
• Subtask 3: 0 / 80