제출 #69853

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
69853		Benq	Sparklers (JOI17_sparklers)	C++14	0 / 100	3 ms	844 KiB

sparklers

#pragma GCC optimize ("O3")
#pragma GCC target ("sse4")

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

using namespace std;
using namespace __gnu_pbds;
using namespace __gnu_cxx;
 
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 N,K,T,mid;
vi X;
ld dp[1000][1000][2];

void mn(ld& a, ld b) { a = min(a,b); }

ld getTi(ld a, ld b) { return (b-a)/2/mid; }

// k = 0: 
// X[i-1]+t*mid 
// currently at X[i]+t*mid 
// X[j]-t*mid 
// X[j+1]-t*mid

void balance(int i, int j) {
    ld t = dp[i][j][0], lef = (ld)(j-i+1)*T-t;
    if (t < INF && lef >= getTi(X[i]+t*mid,X[j]-t*mid)) mn(dp[i][j][1],t+getTi(X[i]+t*mid,X[j]-t*mid));
    
    t = dp[i][j][1], lef = (ld)(j-i+1)*T-t;
    if (t < INF && lef >= getTi(X[i]+t*mid,X[j]-t*mid)) mn(dp[i][j][0],t+getTi(X[i]+t*mid,X[j]-t*mid));
}

void ad(int i, int j) {
    ld t = dp[i][j][0], lef = (ld)(j-i+1)*T-t;
    if (t < INF && i && lef >= getTi(X[i-1],X[i])) mn(dp[i-1][j][0],dp[i][j][0]+getTi(X[i-1],X[i]));
    
    t = dp[i][j][1], lef = (ld)(j-i+1)*T-t;
    if (t < INF && j < N-1 && lef >= getTi(X[j],X[j+1])) mn(dp[i][j+1][1],dp[i][j][1]+getTi(X[j],X[j+1]));
}

bool ok() {
    F0Rd(i,K+1) FOR(j,K,N) F0R(k,2) dp[i][j][k] = INF;
    dp[K][K][0] = dp[K][K][1] = 0;
    F0Rd(i,K+1) FOR(j,K,N) balance(i,j), ad(i,j);
    return min(dp[0][N-1][0],dp[0][N-1][1]) != INF;
}

int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin >> N >> K >> T; X.resize(N); K--;
    F0R(i,N) cin >> X[i];
    int lo = 0, hi = MOD;
    while (lo < hi) {
        mid = (lo+hi)/2;
        if (ok()) hi = mid;
        else lo = mid+1;
    }
    cout << lo;
}

/* Look for:
* the exact constraints (multiple sets are too slow for n=10^6 :( ) 
* special cases (n=1?)
* overflow (ll vs int?)
* array bounds
* if you have no idea just guess the appropriate well-known algo instead of doing nothing :/
*/

• Subtask 1: 0 / 30
• Subtask 2: 0 / 20
• Subtask 3: 0 / 50