Submission #431972

#	Time	Username	Problem	Language	Result	Execution time	Memory
431972		MarcoMeijer	Skyscraper (JOI16_skyscraper)	C++14	100 / 100	422 ms	312968 KiB

skyscraper

#include <bits/stdc++.h>
using namespace std;
 
// macros
typedef long long ll;
typedef long double ld;
typedef pair<int, int> ii;
typedef pair<ll, ll> lll;
typedef tuple<int, int, int> iii;
typedef vector<int> vi;
typedef vector<ii> vii;
typedef vector<iii> viii;
typedef vector<ll> vll;
typedef vector<lll> vlll;
#define REP(a,b,c) for(int a=int(b); a<int(c); a++)
#define RE(a,c) REP(a,0,c)
#define RE1(a,c) REP(a,1,c+1)
#define REI(a,b,c) REP(a,b,c+1)
#define REV(a,b,c) for(int a=int(c-1); a>=int(b); a--)
#define FOR(a,b) for(auto& a : b)
#define all(a) a.begin(), a.end()
#define INF 1e18
#define EPS 1e-9
#define pb push_back
#define popb pop_back
#define fi first
#define se second
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
 
// input
template<class T> void IN(T& x) {cin >> x;}
template<class H, class... T> void IN(H& h, T&... t) {IN(h); IN(t...); }
 
// output
template<class T1, class T2> void OUT(const pair<T1,T2>& x);
template<class T> void OUT(const vector<T>& x);
template<class T> void OUT(const T& x) {cout << x;}
template<class H, class... T> void OUT(const H& h, const T&... t) {OUT(h); OUT(t...); }
template<class T1, class T2> void OUT(const pair<T1,T2>& x) {OUT(x.fi,' ',x.se);}
template<class T> void OUT(const vector<T>& x) {RE(i,x.size()) OUT(i==0?"":" ",x[i]);}
template<class... T> void OUTL(const T&... t) {OUT(t..., "\n"); }
template<class H> void OUTLS(const H& h) {OUTL(h); }
template<class H, class... T> void OUTLS(const H& h, const T&... t) {OUT(h,' '); OUTLS(t...); }
 
// dp
template<class T> bool ckmin(T&a, T&b) { bool bl = a > b; a = min(a,b); return bl;}
template<class T> bool ckmax(T&a, T&b) { bool bl = a < b; a = max(a,b); return bl;}
 
void program();
int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    program();
}
 

// mod library
ll MOD=1e9+7;

inline ll mod(ll x_) {
    return (x_)%MOD;
}
ll modpow(ll x_, ll N_) {
    if(N_ == 0) return 1;
    ll a = modpow(x_,N_/2);
    a = (a*a)%MOD;
    if(N_%2) a = (a*x_)%MOD;
    return a;
}
ll inv(ll x_) {
    return modpow(x_, MOD-2);
}
class mi {
public:
    mi(ll v=0) {value = v;}
    mi  operator+ (ll rs) {return mod(value+rs);}
    mi  operator- (ll rs) {return mod(value-rs+MOD);}
    mi  operator* (ll rs) {return mod(value*rs);}
    mi  operator/ (ll rs) {return mod(value*inv(rs));}
    mi& operator+=(ll rs) {*this = (*this)+rs; return *this;}
    mi& operator-=(ll rs) {*this = (*this)-rs; return *this;}
    mi& operator*=(ll rs) {*this = (*this)*rs; return *this;}
    mi& operator/=(ll rs) {*this = (*this)/rs; return *this;}
    operator ll&() {return value;}

    ll value;
};
typedef vector<mi> vmi;

//===================//
//   begin program   //
//===================//
 
const int MX = 110;
const int N = (1<<20);

int n, l, a[MX];
mi dp[MX][MX][MX*10][3];

void program() {
    IN(n,l);
    if(n == 1) return OUTL(1);
    RE1(i,n) IN(a[i]);
    sort(a+1,a+n+1);
    a[n+1] = 1e5;
    dp[0][0][0][0] = 1;
    RE1(i,n) { // first i buildings
        RE1(j,i) { // j components
            RE(k,l+1) { // total cost
                RE(m,3) { // m end points so far
                    int delta = (2*j - m)*(a[i+1]-a[i]);
                    if(delta > k || i + j + 1 - m > n) continue;

                    mi res = 0;

                    // create new component
                    res += dp[i-1][j-1][k - delta][m];

                    // create now component on endpoint
                    if(m) res += mi(3-m)*dp[i-1][j-1][k - delta][m-1];

                    // append to existing component
                    res += mi(2*j - m)*dp[i-1][j][k-delta][m];

                    // append to existing component on endpoint
                    if(m == 1) res += mi(2*j)*dp[i-1][j][k - delta][m-1];
                    if(m == 2) {
                        if(i == n) res += dp[i-1][j][k - delta][m-1];
                        else res += mi(j - 1)*dp[i-1][j][k - delta][m-1];
                    }

                    // join two existing components
                    if(m == 2) {
                        if(i == n) res += dp[i-1][j+1][k - delta][m];
                        else res += mi(j)*mi(j-1)*dp[i-1][j+1][k - delta][m];
                    }
                    else if(m == 1) res += mi(j)*mi(j)*dp[i-1][j+1][k - delta][m];
                    else res += mi(j)*mi(j+1)*dp[i-1][j+1][k - delta][m];

                    dp[i][j][k][m] = res;
                }
            }
        }
    }

    mi ans = 0;
    RE(k,l+1) ans += dp[n][1][k][2];
    OUTL(ll(ans));
}

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