Submission #59035

#	Time	Username	Problem	Language	Result	Execution time	Memory
59035		Crown	Shortcut (IOI16_shortcut)	C++14	31 / 100	2033 ms	1640 KiB

shortcut

#include "shortcut.h"

#include <bits/stdc++.h>
using namespace std;
#define X first
#define Y second
#define pb push_back
typedef pair<int, int> ii;
typedef long long ll;

const int maxn = 505;
vector<ll> qs;
vector<int> l, d;
int c, n;
long long find_shortcut(int _n, vector<int> _l, vector<int> _d, int _c)
{
	l = _l; d = _d; c = _c; n = _n;
	ll ret = 1e18;
	for(int a = 0; a< n; a++)
	{
		for(int b = a+1; b< n; b++)
		{
			//printf("oh %d and %d is\n", a, b);
			ll res = -1e18;
			vector<ll> L, D;
			for(int i = a; i< b; i++) L.pb(l[i]);
			L.pb(c);
			int sz = b-a+1;
			D.assign(sz, 0);
			for(int i = a+1; i< b; i++)
			{
				D[i-a] = d[i];
			}
			ll run = 0;
			ll best_run = 0;
			for(int i = 0; i<= a; i++)
			{
				res = max(res, best_run + run + d[i]);
				best_run = max(best_run, d[i]-run);
				run += l[i];
			}
			best_run = 0;
			run = 0;
			for(int i = b; i< n; i++)
			{  
				res = max(res, best_run+run+d[i]);
				best_run = max(best_run, d[i]-run);
				if(i+1< n) run += l[i];
			}
			//printf("not in cycle %lld\n", res);
			run = 0;
			for(int i = a; i>= 0; i--)
			{
				D[0] = max(D[0], run+d[i]);
				if(i) run += l[i-1];
			}
			run = 0;
			for(int i = b; i< n; i++)
			{
				D.back() = max(D.back(), run+d[i]);
				if(i< n-1) run += l[i];
			}
			for(int i = 0; i< sz; i++) D.pb(D[i]);
			vector< ll > Q;
			for(int i = 0; i< sz; i++)
			{
				if(Q.empty()) Q.pb(L[i]);
				else Q.pb(Q.back()+L[i]);
			}
			ll tot = Q.back();
			//printf("tot = %lld\n", tot);
			for(int i = 0; i+1< sz; i++)
			{
				Q.pb(Q.back()+L[i]);
			}
			//for(auto x : D) printf("%lld ", x);
			//printf("\n");
			int lim = 0;
			deque< pair<long long, int> > dq;
			for(int i = 0; i< sz+sz; i++)
			{
				while(i && 2LL*(Q[i-1]-(lim?Q[lim-1]:0))> tot) lim++;
				//printf("i = %d, lim = %d\n", i, lim);
				while(!dq.empty() && dq.front().Y< lim) dq.pop_front();
				if(!dq.empty()) 
				{
					res = max(res, D[i]+Q[i-1]+dq.front().X);
				}
				while(!dq.empty() && dq.back().X< (D[i]-(i?Q[i-1]:0))) dq.pop_back();
				dq.push_back(make_pair(D[i]-(i?Q[i-1]:0), i));
			}
			//printf("res is %lld\n", res);
			ret = min(ret, res);
		}
	}
	return ret;
}

• Subtask 1: 9 / 9
• Subtask 2: 14 / 14
• Subtask 3: 8 / 8
• Subtask 4: 0 / 7
• Subtask 5: 0 / 33
• Subtask 6: 0 / 22
• Subtask 7: 0 / 4
• Subtask 8: 0 / 3