Submission #60331

#	Time	Username	Problem	Language	Result	Execution time	Memory
60331		Benq	Skyscraper (JOI16_skyscraper)	C++11	100 / 100	717 ms	3212 KiB

skyscraper

#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>

using namespace std;
using namespace __gnu_pbds;
 
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;

typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;

typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;

template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;

#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define F0R(i, a) for (int i=0; i<(a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)

#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()

const int MOD = 1000000007;
const ll INF = 1e18;
const int MX = 100001;

int mul(int a, int b) { return (ll)a*b%MOD; }
int ad(int a, int b) { return (a+b)%MOD; }

void MUL(int& a, int b) { a = mul(a,b); }
void AD(int& a, int b) { a = ad(a,b); }
void MN(int& a, int b) { a = min(a,b); }

int N,L, mn[3][101], mnTMP[3][101]; 
int dp[3][101][1001], dpTMP[3][101][1001];
vi A;

void tri(int a, int b, int c, int d) {
    if (c > mnTMP[a][b]+L) return;
    AD(dpTMP[a][b][c-mnTMP[a][b]],d);
}

void triEnd(int a, int b, int c, int x) {
    if (!a) return;
    int val = mul(a,dp[a][b][c]);
    c += mn[a][b]+x; a --;
    F0R(i,2) F0R(j,2) tri(a+i,b+j,c-i*x-2*j*x,val); 
}

void triMid(int a, int b, int c, int x) {
    if (!b) return;
    int val = mul(b,dp[a][b][c]);
    c += mn[a][b]+2*x; b --;
    F0R(i,2) F0R(j,2) tri(a,b+i+j,c-2*i*x-2*j*x,val);
}

void process(int x) {
    F0R(i,3) F0R(j,N+1) {
        mnTMP[i][j] = MOD;
        F0R(k,L+1) dpTMP[i][j][k] = 0;
    }
    
    F0R(i,3) F0R(j,N+1) if (mn[i][j] != MOD) {
        if (i) F0R(I,2) F0R(J,2) MN(mnTMP[i-1+I][j+J],mn[i][j]+x-I*x-2*J*x);
        if (j) F0R(I,2) F0R(J,2) MN(mnTMP[i][j-1+I+J],mn[i][j]+2*x-2*I*x-2*J*x);
    }

    F0R(i,3) F0R(j,N+1) if (mn[i][j] != MOD) F0R(k,L+1) {
        triEnd(i,j,k,x);
        triMid(i,j,k,x);
    }
    
    F0R(i,3) F0R(j,N+1) {
        mn[i][j] = mnTMP[i][j];
        F0R(k,L+1) dp[i][j][k] = dpTMP[i][j][k];
    }
    
    // cout << "AH " << mn[0][0] << "\n";
}

void finish() {
    int ans = 0;
    // cout << "ZZ " << mn[0][0] << "\n";
    F0R(i,L-mn[0][0]+1) AD(ans,dp[0][0][i]);
    cout << ans;
}

void input() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin >> N >> L; A.resize(N);
    F0R(i,N) cin >> A[i];
    sort(all(A));
    //cout << A[0] << " " << A[N-1] << "\n";
}

void init(int x) {
    F0R(i,3) F0R(j,N+1) mn[i][j] = MOD;
    F0R(i,2) F0R(j,2) MN(mn[i+j][0],-(i+j)*A[0]);
    F0R(i,2) F0R(j,2) dp[i+j][0][0] ++;
}

int main() {
    input();
    init(A[0]);
    FOR(i,1,sz(A)) process(A[i]);
    finish();
}

/* Look for:
* the exact constraints (multiple sets are too slow for n=10^6 :( ) 
* special cases (n=1?)
* overflow (ll vs int?)
* array bounds
*/

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