Submission #642476

#	Time	Username	Problem	Language	Result	Execution time	Memory
642476		BackNoob	Watching (JOI13_watching)	C++14	100 / 100	236 ms	16212 KiB

watching

#include <bits/stdc++.h>
#define ll long long
#define fi first
#define se second
#define endl '\n'
#define MASK(i) (1LL << (i))
#define ull unsigned long long
#define ld long double
#define pb push_back
#define all(x) (x).begin() , (x).end()
#define BIT(x , i) ((x >> (i)) & 1)
#define TASK "task"
#define sz(s) (int) (s).size()
 
using namespace std;
const int mxN = 2007;
const int inf = 1e9 + 277;
const int mod = 1e9 + 7;
const ll infll = 1e18 + 7;
const int base = 307;
const int LOG = 20;
 
template <typename T1, typename T2> bool minimize(T1 &a, T2 b) {
    if (a > b) {a = b; return true;} return false;
}
template <typename T1, typename T2> bool maximize(T1 &a, T2 b) {
    if (a < b) {a = b; return true;} return false;
}
 
int n;
int p;
int q;
int a[mxN];
int nxt1[mxN];
int nxt2[mxN];
int dp[mxN][mxN];

bool ok(int x)
{
    int j1 = 1;
    int j2 = 1;
    for(int i = 1; i <= n; i++) {
        j1 = max(j1, i);
        j2 = max(j2, i);
        while(j1 < n && a[j1 + 1] - a[i] + 1 <= x) ++j1;
        while(j2 < n && a[j2 + 1] - a[i] + 1 <= 2 * x) ++j2;

        nxt1[i] = j1;
        nxt2[i] = j2;
    }


    memset(dp, 0x3f, sizeof(dp));
    int tmp = dp[0][0];
    dp[0][0] = 0;
    for(int i = 0; i < n; i++) {
        for(int j = 0; j <= p; j++) {
            if(dp[i][j] == tmp) continue;
            minimize(dp[nxt1[i + 1]][j + 1], dp[i][j]);
            minimize(dp[nxt2[i + 1]][j], dp[i][j] + 1);
        }
    }

    for(int j = 0; j <= p; j++) {
        if(dp[n][j] <= q) return true;
    }
    return false;
}
 
void solve()
{
    cin >> n >> p >> q;
    for(int i = 1; i <= n; i++) cin >> a[i];

    p = min(p, n);
    q = min(q, n);

    sort(a + 1, a + 1 + n);
    int lo = 0, hi = inf;
    while(lo + 1 < hi) {
        int mid = (lo + hi) / 2;
        if(ok(mid)) hi = mid;
        else lo = mid;
    }

    cout << hi;
}
 
int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);

 
    int tc = 1;
    //cin >> tc;
    while(tc--) {
        solve();
    }
    return 0;
}

• Subtask 1: 50 / 50
• Subtask 2: 50 / 50