Submission #1051804

#	Time	Username	Problem	Language	Result	Execution time	Memory
1051804		ReLice	Aliens (IOI16_aliens)	C++17	4 / 100	16 ms	125952 KiB

aliens

#include "aliens.h"
#include <bits/stdc++.h>
#define ll long long
#define sz size()
#define all(x) x.begin(),x.end()
#define vll vector<ll>
#define str string
#define pb push_back
#define pof pop_front()
#define pf push_front
#define pob pop_back()
#define bc back()
using namespace std;
const ll N=4e3+5;
struct Line{
	ll m,b;
	
	ll operator () (ll x){return m * x + b;}
	
	bool ok(Line x, Line y){
		ll t1 = (b - x.b);
		ll b1 = (x.m - m);
		ll t2 = (x.b - y.b);
		ll b2 = (y.m - x.m);
		return t1 * b2 <= t2 * b1;
	}
	
};

ll sq(ll x){
	return x*x;
}
long long dp[N][N];
long long take_photos(int N, int M, int K, vector<int> R, vector<int> C) {
	ll i,q;
	ll n=N,m=M,k=K;
	vll len, nxt;
	vll c, r;
	for(auto i : C)c.pb(i);
	for(auto i : R)r.pb(i);
	
	
	{//deleting unnecessary ones
		vll mx(m, m), nr;
		
		for(i=0;i<n;i++){
			if(r[i] < c[i]) swap(r[i], c[i]);
			
			mx[r[i]] = min(mx[r[i]], c[i]);
		} 
		for(i=m-1;i>=0;i--){
			
			if(mx[i] > i)continue;
			
			if(nr.empty() || mx[nr.bc] > mx[i]){
				nr.pb(i);
				len.pb(i-mx[i]+1);
			}
		}
		
		nr.pb(-1), len.pb(-1);//changing indexation

		reverse(all(nr));
		reverse(all(len));
		
		r=nr;
		n=r.size()-1;
		k = min(k, n);
	}
	
	c.clear();
	c.pb(0);
	for(i=1;i<=n;i++){
		nxt.pb(r[i] - len[i]);
		c.pb(nxt.bc);
		len[i - 1] = r[i - 1] - nxt[i - 1] ;
		len[i - 1] = max(len[i - 1], 0ll);
	}
	len[0]=0;
	nxt.pb(r[n]);
	
	memset(dp, 0x3f, sizeof(dp));
	dp[0][1]=0;
	
	vector<deque<Line>> ch(k+5);
	
	for(i=0;i<=k;i++)ch[i].pb({-2 * nxt[0], nxt[0] * nxt[0]});
	
	for(i=1;i<=n;i++){ 
		for(q=1;q<=min(k,i+1);q++){
			auto &dq = ch[q-1];
			ll x = r[i];
			while(dq.sz > 1 && dq[0](x) > dq[1](x)){
				dq.pof;
			}
			if(dq.sz){
				dp[i][q] = r[i] * r[i] + dq[0](x);	
			}
		}
		
		for(q=1;q<=k;q++){
			if(dp[i][q] >= dp[n+1][0]) continue;
			ll m = (-2) * nxt[i];
			ll b = nxt[i] * nxt[i] - len[i] * len[i] + dp[i][q];
			Line cur = {m, b};
			auto &dq = ch[q];
			while((ll) dq.sz > 1 && dq[dq.sz - 2].ok(dq.bc, cur)) dq.pob;
			dq.pb(cur); 
		}
	}
	
	long long ans=dp[n+1][0];
	for(i=0;i<=k;i++) ans=min(ans,dp[n][i]);
    
    return ans;
}

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