Submission #389926

#	Time	Username	Problem	Language	Result	Execution time	Memory
389926		smax	Skyscraper (JOI16_skyscraper)	C++17	100 / 100	351 ms	260596 KiB

skyscraper

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

#ifdef LOCAL
#define DEBUG(...) debug(#__VA_ARGS__, __VA_ARGS__)
#else
#define DEBUG(...) 6
#endif

template<typename T, typename S> ostream& operator << (ostream &os, const pair<T, S> &p) {return os << "(" << p.first << ", " << p.second << ")";}
template<typename C, typename T = decay<decltype(*begin(declval<C>()))>, typename enable_if<!is_same<C, string>::value>::type* = nullptr>
ostream& operator << (ostream &os, const C &c) {bool f = true; os << "["; for (const auto &x : c) {if (!f) os << ", "; f = false; os << x;} return os << "]";}
template<typename T> void debug(string s, T x) {cerr << s << " = " << x << "\n";}
template <typename T, typename... Args> void debug(string s, T x, Args... args) {for (int i=0, b=0; i<(int)s.size(); i++) if (s[i] == '(' || s[i] == '{') b++; else
if (s[i] == ')' || s[i] == '}') b--; else if (s[i] == ',' && b == 0) {cerr << s.substr(0, i) << " = " << x << " | "; debug(s.substr(s.find_first_not_of(' ', i + 1)), args...); break;}}

template<int MOD>
struct ModInt {
    long long v;
    ModInt(long long _v = 0) {v = (-MOD < _v && _v < MOD) ? _v : _v % MOD; if (v < 0) v += MOD;}
    ModInt& operator += (const ModInt &other) {v += other.v; if (v >= MOD) v -= MOD; return *this;}
    ModInt& operator -= (const ModInt &other) {v -= other.v; if (v < 0) v += MOD; return *this;}
    ModInt& operator *= (const ModInt &other) {v = v * other.v % MOD; return *this;}
    ModInt& operator /= (const ModInt &other) {return *this *= inverse(other);}
    bool operator == (const ModInt &other) const {return v == other.v;}
    bool operator != (const ModInt &other) const {return v != other.v;}
    friend ModInt operator + (ModInt a, const ModInt &b) {return a += b;}
    friend ModInt operator - (ModInt a, const ModInt &b) {return a -= b;}
    friend ModInt operator * (ModInt a, const ModInt &b) {return a *= b;}
    friend ModInt operator / (ModInt a, const ModInt &b) {return a /= b;}
    friend ModInt operator - (const ModInt &a) {return 0 - a;}
    friend ModInt power(ModInt a, long long b) {ModInt ret(1); while (b > 0) {if (b & 1) ret *= a; a *= a; b >>= 1;} return ret;}
    friend ModInt inverse(ModInt a) {return power(a, MOD - 2);}
    friend ostream& operator << (ostream &os, const ModInt &m) {return os << m.v;}
};
typedef ModInt<1000000007> M;

// number of placed values, number of components, sum of abs(f_i - f_{i+1}), number of end gaps filled
M dp[105][105][1005][3];

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    int n, l;
    cin >> n >> l;
    vector<int> a(n + 1);
    for (int i=0; i<n; i++)
        cin >> a[i];
    a[n] = 1e4;

    if (n == 1) {
        cout << "1\n";
        return 0;
    }

    sort(a.begin(), a.end());
    // form new individual component, joined to one of ends
    if (a[1] - a[0] <= l)
        dp[1][1][a[1] - a[0]][1] = 2;
    // form new individual component, not joined to end
    if (2 * (a[1] - a[0]) <= l)
        dp[1][1][2 * (a[1] - a[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<3; m++) {
                    int diff = a[i+1] - a[i];
                    // joined to one of ends
                    if (m < 2 && k + (2 * j - m + 1) * diff <= l)
                        dp[i+1][j+1][k+(2*j-m+1)*diff][m+1] += dp[i][j][k][m] * (2 - m);
                    // joined to an existing component
                    if (k + (2 * j - m) * diff <= l)
                        dp[i+1][j][k+(2*j-m)*diff][m] += dp[i][j][k][m] * (2 * j - m);
                    // merge two existing components together
                    if (j > 1 && k + (2 * (j - 1) - m) * diff <= l)
                        dp[i+1][j-1][k+(2*(j-1)-m)*diff][m] += dp[i][j][k][m] * (j - 1);
                    // merge a component with an end
                    if (m < 2 && k + (2 * j - m - 1) * diff <= l)
                        dp[i+1][j][k+(2*j-m-1)*diff][m+1] += dp[i][j][k][m] * (2 - m);
                    // not joined to anything
                    if (k + (2 * (j + 1) - m) * diff <= l)
                        dp[i+1][j+1][k+(2*(j+1)-m)*diff][m] += dp[i][j][k][m] * (j + 1 - m);
                }

    M ret = 0;
    for (int k=0; k<=l; k++)
        ret += dp[n][1][k][2];
    cout << ret << "\n";

    return 0;
}

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