DP- Twix

Pitanja i problemi
1

Pitanje ili opis problema

Imam problem sa zadatkom Twix iz dinamickog programiranja. Imao sam zanmljivu ideju da sabiram poredim ukupne sume izmedju dva reda i da dodam onaj element koji pripada delu sa manjom sumom, ali samo dva test primera prolaze. Da li uopste ova ideja moze da radi ili bi mi znacila neka druga bolja ideja?

Ovo je moj kod:

#include<bits/stdc++.h>

using namespace std;

int n;
long long s[10000010][2];
long long  p[10000010][2];

int main(){
    std::cin.sync_with_stdio(false);
    cin>>n;
    cin>>s[1][0]>>s[1][1];
    for(int i=2; i<=n; i++){
        long long  x, y;
        cin>>x>>y;
        s[i][0]= s[i-1][0]+x;
        s[i][1]= s[i-1][1]+y;
        cin>>p[i][0]>>p[i][1];
    }


    long long sum=0;
    int curr;

    if(s[n][0]< s[n][1]){
        sum+= s[1][0];
        curr=0;
    }
    else{
        sum+= s[1][1];
        curr=1;
    }

    //cout<<sum<<"\n";

    for(int i=2; i<=n; i++){
        if((s[n][!curr]-s[i-1][!curr]+ p[i][curr]) < (s[n][curr]- s[i-1][curr])){
            sum+=p[i][curr];
            curr=!curr;
        }
        sum+= s[i][curr]- s[i-1][curr];
        //cout<<sum<<"\n";

    }

    cout<<sum<<"\n";

    return 0;
}

Link ka zadatku ili odgovarajućoj stranici

https://petlja.org/biblioteka/r/Problems/Twix

2

Recimo da je dp[i][0] minimalno vreme tako da kolac u i-tom trenutku bude u prvoj fabrici, dok je dp[i][1] minimalno vreme tako da kolac bude u drugoj fabrici u i-tom trenutku. Sada lako moze da se dobije formula prelaza: dp[i][0] = min(dp[i - 1][0] + S1[i], dp[i - 1][1] + S1[i] + P21[i]), dok je dp[i][1] = min(dp[i - 1][1] + S2[i], dp[i - 1][0] + S2[i] + P12[i]).
Evo ga i kod:

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef long double ld;
typedef double db;
typedef string str;

typedef pair<int,int> pi;
typedef pair<ll,ll> pl;
typedef pair<db,db> pd;

typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<db> vd;
typedef vector<str> vs;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<pd> vpd;

#define mp make_pair
#define sz(x) (int)(x).size()
#define all(x) begin(x), end(x)
#define rall(x) (x).rbegin(), (x).rend()
#define rsz resize
#define ins insert
#define ft front()
#define bk back()
#define pf push_front
#define pb push_back
#define eb emplace_back
#define lb lower_bound
#define ub upper_bound

#define FOR(i,a,b) for (int i = (a); i < (b); ++i)
#define F0R(i,a) FOR(i,0,a)
#define ROF(i,a,b) for (int i = (b)-1; i >= (a); --i)
#define R0F(i,a) ROF(i,0,a)
#define trav(a,x) for (auto& a: x)

template<class A> void re(complex<A>& c);
template<class A, class B> void re(pair<A,B>& p);
template<class A> void re(vector<A>& v);
template<class A, size_t SZ> void re(array<A,SZ>& a);

template<class T> void re(T& x) { cin >> x; }
void re(db& d) { str t; re(t); d = stod(t); }
void re(ld& d) { str t; re(t); d = stold(t); }
template<class H, class... T> void re(H& h, T&... t) { re(h); re(t...); }

template<class A> void re(complex<A>& c) { A a,b; re(a,b); c = {a,b}; }
template<class A, class B> void re(pair<A,B>& p) { re(p.f,p.s); }
template<class A> void re(vector<A>& x) { trav(a,x) re(a); }
template<class A, size_t SZ> void re(array<A,SZ>& x) { trav(a,x) re(a); }

#define ts to_string
str ts(char c) { return str(1,c); }
str ts(const char* s) { return (str)s; }
str ts(str s) { return s; }
str ts(bool b) {
	#ifdef LOCAL
		return b ? "true" : "false";
	#else
		return ts((int)b);
	#endif
}
template<class A> str ts(complex<A> c) {
	stringstream ss; ss << c; return ss.str(); }
str ts(vector<bool> v) {
	str res = "{"; F0R(i,sz(v)) res += char('0'+v[i]);
	res += "}"; return res; }
template<size_t SZ> str ts(bitset<SZ> b) {
	str res = ""; F0R(i,SZ) res += char('0'+b[i]);
	return res; }
template<class A, class B> str ts(pair<A,B> p);
template<class T> str ts(T v) { // containers with begin(), end()
	#ifdef LOCAL
		bool fst = 1; str res = "{";
		for (const auto& x: v) {
			if (!fst) res += ", ";
			fst = 0; res += ts(x);
		}
		res += "}"; return res;
	#else
		bool fst = 1; str res = "";
		for (const auto& x: v) {
			if (!fst) res += " ";
			fst = 0; res += ts(x);
		}
		return res;

	#endif
}
template<class A, class B> str ts(pair<A,B> p) {
	#ifdef LOCAL
		return "("+ts(p.f)+", "+ts(p.s)+")";
	#else
		return ts(p.f)+" "+ts(p.s);
	#endif
}

template<class A> void pr(A x) { cout << ts(x); }
template<class H, class... T> void pr(const H& h, const T&... t) {
	pr(h); pr(t...); }
void ps() { pr("\n"); } // print w/ spaces
template<class H, class... T> void ps(const H& h, const T&... t) {
	pr(h); if (sizeof...(t)) pr(" "); ps(t...); }

const int MOD = 1e9+7;
const int MX = 2e5;
const int INF = INT_MAX;
mt19937 rng((uint32_t)chrono::steady_clock::now().time_since_epoch().count());

void ckmax(int& a,int b){a=max(a,b);}
void ckmax(ll& a,ll b){a=max(a,b);}
void ckmin(int& a,int b){a=min(a,b);}
void ckmin(ll& a,ll b){a=min(a,b);}

int main(){
    int N; re(N);
    vi S1(N),S2(N);
    re(S1[0]); re(S2[0]);
    vi change1(N),change2(N);
    FOR(i,1,N) re(S1[i],S2[i],change1[i],change2[i]);
    vector<vi> dp(N,vi(2,INF));
    dp[0][0]=S1[0];
    dp[0][1]=S2[0];
    FOR(i,1,N){
 		dp[i][0]=min(dp[i-1][0]+S1[i],dp[i-1][1]+S1[i]+change2[i]);
 		dp[i][1]=min(dp[i-1][1]+S2[i],dp[i-1][0]+S2[i]+change1[i]);
    }
    pr(min(dp[N-1][0],dp[N-1][1]));
    return 0;
}
3

Hvala puno, razumeo sam :smiley:

1 Like