Submission #1036780

#	Time	Username	Problem	Language	Result	Execution time	Memory
1036780		c2zi6	Aliens (IOI16_aliens)	C++14	41 / 100	2070 ms	94312 KiB

aliens

#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
#define ff first
#define ss second
#define pb push_back
#define all(a) (a).begin(), (a).end()
#define replr(i, a, b) for (int i = int(a); i <= int(b); ++i)
#define reprl(i, a, b) for (int i = int(a); i >= int(b); --i)
#define rep(i, n) for (int i = 0; i < int(n); ++i)
#define mkp(a, b) make_pair(a, b)
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef vector<PII> VPI;
typedef vector<VI> VVI;
typedef vector<VVI> VVVI;
typedef vector<VPI> VVPI;
typedef pair<ll, ll> PLL;
typedef vector<ll> VL;
typedef vector<PLL> VPL;
typedef vector<VL> VVL;
typedef vector<VVL> VVVL;
typedef vector<VPL> VVPL;
template<class T> T setmax(T& a, T b) {if (a < b) return a = b; return a;}
template<class T> T setmin(T& a, T b) {if (a < b) return a; return a = b;}
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
template<class T>
using indset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
#include "aliens.h"

ll square(ll a) {
    return a*a;
}

ll take_photos(int n, int m, int k, VI R, VI C) {
    VPI pts;
    rep(i, n) {
        int r = R[i];
        int c = C[i];
        if (r > c) swap(r, c);
        pts.pb({r, c});
    }
    {
        VPI tmp;
        tmp.pb({-1, -1});
        sort(all(pts));
        rep(i, n) {
            if (i == n-1 || pts[i+1].ff != pts[i].ff) {
                if (!tmp.size() || tmp.back().ss < pts[i].ss) {
                    tmp.pb(pts[i]);
                }
            }
        }
        pts = tmp;
        n = pts.size()-1;
    }

    auto cost = [&](int j, int i) {
        return square(pts[i].ss - pts[j+1].ff + 1) - square(max(0, pts[j].ss - pts[j+1].ff + 1));
    };

    VVL dp(n+1, VL(k+1, 1e18));
    VVI opt(n+1, VI(k+1));
    replr(j, 0, k) dp[0][j] = 0;
    /*replr(i, 1, n) {*/
    /*    replr(j, 1, k) {*/
    /*        replr(l, 0, i-1) {*/
    /*            ll cur = dp[l][j-1] + cost(l, i);*/
    /*            if (cur < dp[i][j]) {*/
    /*                dp[i][j] = cur;*/
    /*                opt[i][j] = l;*/
    /*            }*/
    /*        }*/
    /*    }*/
    /*}*/
    replr(D, 1-k, n-1) replr(i, 1, n) {
        int j = i-D;
        if (j < 1 || j > k) continue;
        int mn = 0;
        int mx = i-1;
        if (i-1 >= 1) mn = opt[i-1][j];
        if (j+1 <= k) mx = opt[i][j+1];
        replr(l, mn, mx) {
            ll cur = dp[l][j-1] + cost(l, i);
            if (cur < dp[i][j]) {
                dp[i][j] = cur;
                opt[i][j] = l;
            }
        }
    }
    return dp[n][k];
}

• Subtask 1: 4 / 4
• Subtask 2: 12 / 12
• Subtask 3: 9 / 9
• Subtask 4: 16 / 16
• Subtask 5: 0 / 19
• Subtask 6: 0 / 40