POJ3624 01背包问题

Charm Bracelet

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 30427		Accepted: 13581

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 W_i (1 ≤ W_i ≤ 400), a 'desirability' factor D_i (1 ≤ D_i ≤ 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
* Lines 2..N+1: Line i+1 describes charm i with two space-separated integers: W_i and D_i

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

经典的背包问题。

遇到的问题和思路：

       本想用v[i][j]的二维数组来做的，结果13000 * 3500 >1600W了，于是想到用一维的数组反复操作即可

#include<cstdio> #include<algorithm> #include<cstring> using namespace std; const int maxn = 3000 + 500; int n, m; int w[maxn], d[maxn]; int v[13000]; int main(){ while(scanf("%d%d", &n, &m) != EOF){ memset(w, 0, sizeof(w)); memset(v, 0, sizeof(v)); memset(d, 0, sizeof(d)); for (int i = 1; i <= n; i++){ scanf("%d%d", &w[i], &d[i]); } for (int i = 1; i <= n; i++){ for (int j = m; j >= w[i]; j--){ v[j] = max(v[j], v[j - w[i]] + d[i]); } } printf("%d\n", v[m]); } return 0; } </cstring></algorithm></cstdio>