Cod sursa(job #2191483)

Utilizator codrutciulacuciulacu codrut cristian codrutciulacu Data 2 aprilie 2018 21:19:32
Problema Problema rucsacului Scor 25
Compilator cpp Status done
Runda Arhiva educationala Marime 1.66 kb
/*#include <cstdio>
#include <algorithm>

#define NMAX 5010
#define GMAX 10010

int M[NMAX][GMAX];
int W[NMAX], P[NMAX];

void creare_matrice(int i, int cw){
	M[i][cw] = M[i - 1][cw];

	if(cw >= W[i])
		M[i][cw] = std::max(M[i - 1][cw], M[i - 1][cw - W[i]] + P[i]);
}

int main(){
	register int N;
	register int G;
	freopen("rucsac.in", "r", stdin);
	freopen("rucsac.out", "w", stdout);

	scanf("%d%d", &N, &G);
	for(int i = 1; i <= N; ++i)
		scanf("%d%d", &W[i], &P[i]);


	for(int i = 1; i <= N; ++i){
		for(int cw = 1; cw <= G; ++cw){
			creare_matrice(i, cw);
		}
	}

	printf("%d", M[N][G]);

	return 0;	
}*/
#include <cstdio>
#include <algorithm>
 
using namespace std;
 
#define MAXN 5010
#define MAXG 10010
 
int N, G, Pmax;
int W[MAXN], P[MAXN];
int D[MAXN][MAXG];
 
int main()
{
    freopen("rucsac.in", "r", stdin);
    freopen("rucsac.out", "w", stdout);
 
    // Citire
    scanf("%d%d", &N, &G);
    for(int i = 1; i <= N; ++i)
        scanf("%d%d", &W[i], &P[i]);
 
    // Dinamica D[i][cw] - profitul maxim pe care-l putem obtine adaugand o submultime a primelor i obiecte, insumand greutatea cw
    for(int i = 1; i <= N; ++i)
        for(int cw = 0; cw <= G; ++cw)
        {
            // Mai intai nu punem obiectul i.
            D[i][cw] = D[i-1][cw];
 
            // Daca acest lucru duce la o solutie curenta mai buna, adaugam obiectul i la o solutie anterioara.
            if(W[i] <= cw)
                D[i][cw] = max(D[i][cw], D[i - 1][cw - W[i]] + P[i]);
        }
 
    // Solutia se va afla in statea D[N][G]
    Pmax = D[N][G];
 
    // Afisare
    printf("%d\n", Pmax);
 
    return 0;
}