POJ 3624 Charm Bracelet

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 weightW_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 inte

gers: 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

典型的动态规划问题————0-1背包

dp[i] 表示表示前i件物品放入一个容量为v的背包可以获得的最大价值
因此 状态转移方程可得：

dp[j] = dp[j] > (dp[j-d[i]] + w[i]) ? dp[j] : (dp[j-d[i]] + w[i]);

 1 #include<stdio.h>

 2 #include<string.h>

 3 #define N 12888

 4 

 5 int n, m, d[N],w[N], dp[N];

 6 int main()

 7 {

 8     int i, j;

 9     scanf("%d %d", &n, &m);

10     for(i = 0; i < n; i++)

11         scanf("%d %d", &d[i], &w[i]);

12 

13     memset(dp, 0, sizeof(dp));

14     for(i = 0; i < n; i++)

15         for(j = m; j >= d[i]; j--)

16             dp[j] = dp[j] > (dp[j-d[i]] + w[i]) ? dp[j] : (dp[j-d[i]] + w[i]);

17 

18     printf("%d\n", dp[m]);

19 }