Description Bessie has gone to the mall's jewelry store and spies a charm bracelet. Of course, she'd like to fill it with the best charms possible from the N (1 ≤ N ≤ 3,402) available charms. Each charm i in the supplied list has a weight Wi (1 ≤ Wi≤ 400), a 'desirability' factor Di (1 ≤ Di ≤ 100), and can be used at most once. Bessie can only support a charm bracelet whose weight is no more than M (1 ≤ M ≤ 12,880). Given that weight limit as a constraint and a list of the charms with their weights and desirability rating, deduce the maximum possible sum of ratings. Input * Line 1: Two space-separated integers: N and M Output * Line 1: A single integer that is the greatest sum of charm desirabilities that can be achieved given the weight constraints Sample Input 4 6 1 4 2 6 3 12 2 7 Sample Output 23 Source |
#include <cstdio> #include <cmath> #include <iostream> #include <cstring> #include <algorithm> using namespace std; typedef __int64 Int; typedef long long LL; const double ESP = 1e-5; const double Pi = acos(-1.0); const int MAXN = 40000 + 10; const int INF = 0x3f3f3f3f; int dp[MAXN], N, M, w[MAXN], d[MAXN]; int main() { while (scanf("%d%d", &N, &M) != EOF) { for (int i = 1; i <= N; i++) { scanf("%d%d", &w[i], &d[i]); } memset(dp, 0, sizeof(dp)); for (int i = 1; i <= N; i++) { for (int j = M; j >= w[i]; j--) { dp[j] = max(dp[j], dp[j - w[i]] + d[i]); } } printf("%d ", dp[M]); } return 0; }