Cod sursa(job #2470647)

Utilizator minecraft3Vintila Valentin Ioan minecraft3 Data 9 octombrie 2019 17:21:31
Problema Energii Scor 5
Compilator cpp-64 Status done
Runda Arhiva de probleme Marime 2.3 kb
#include <bits/stdc++.h>
using namespace std;

#define LL long long
#define LD long double
#define PL pair<int,int>
#define PLL pair<long,long>
#define PD  pair<double,double>
#define PLD pair<long double,long double>

#define SORT_PFS(i,j,s1,s2) sort(i,j,[](auto&a,auto&b){return(a.first==b.first)?(a.second s2 b.second):(a.first s1 b.first);})
#define SORT_PSF(i,j,s1,s2) sort(i,j,[](auto&a,auto&b){return(a.second==b.second)?(a.first s2 b.first):(a.second s1 b.second);})
#define CMP(c) [](auto&a, auto&b){return(c);}

template<class T>inline pair<T,T>& swap(pair<T,T>&p){swap(p.first,p.second);return p;}

template<class T1,class T2>ostream&operator<<(ostream&os,pair<T1,T2>p){os<<p.first<<' '<<p.second;return os;}
template<class T1,class T2>ostream&operator>>(ostream&os,pair<T1,T2>p){os<<p.second<<' '<<p.first;return os;}
template<class T1,class T2>istream&operator>>(istream&is,pair<T1,T2>&p){is>>p.first>>p.second;return is;}
template<class T1,class T2>istream&operator<<(istream&is,pair<T1,T2>&p){is>>p.second>>p.first;return is;}

inline LL ipow(LL a,LL b){return(!b)?1:((b%2==0)?1:a)*ipow(a*a,b/2);}
inline LL ipow(LL a,LL b,LL m){return(!b)?1:((b%2==0)?1:a)*ipow(a*a%m,b/2, m)%m;}
inline LL ilog2(LL x){return (!x)?0:1+ilog2(x/2);}

#define pi 3.14159265358979323846
#define MIN(t) numeric_limits< t >::min()
#define MAX(t) numeric_limits< t >::max()

#define FOR(a,b) for(int a=0;a<b;++a)
#define FOR1(a,b) for(int a=1;a<=b;++a)
#define FORD(a,b) for(int a=b-1;a>=0;--a)
#define FORD1(a,b) for(int a=b;a>0;--a)

template<class Tv=int>void read_v(int&n,vector<Tv>&v){cin>>n;v.reserve(n);FOR(i,n)cin>>v[i];}

#define TEST freopen("test.in","r",stdin)
#define FIN(a) ifstream fin(a)
#define FOUT(a) ofstream fout(a)
#define FIO(a) ifstream fin(string(a)+".in");ofstream fout(string(a)+".out")

int main() {
    FIO("energii");
    LL n, cap, g[1005], p[1005], l = (LL)MAX(int) * 100000, s = 0, k;
    LL profit[5005];
    
    fin >> n >> cap;
    FOR(i, n) fin >> p[i] >> g[i], s += p[i];
    if(s < cap) { fout << -1; return 0; }
    FOR1(i, s) profit[i] = l; profit[0] = 0;

    FOR(i, n) { FORD(j, cap)
        if(profit[j] != l) {
            k = (j+p[i] >= cap) ? cap : j + p[i];
            profit[k] = min(profit[k], profit[j] + g[i]);
        }
    }
    fout << profit[cap];
}