POJ 3624 Charm Bracelet （01）（背包入门）

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

01背包主要关系：

if (dp[j]<dp[j-w[i]]+d[i])
            dp[j]=dp[j-w[i]]+d[i];

对于题给的例子：

             1           2        3          2


dp[1]       4           4         4        4   

 
dp[2]       4           6         6        7


dp[3]       4           10       12        12


dp[4]       4           10       16        13



dp[5]       4           10       18        19    
 


dp[6]       4           10       22         23   


dp[j]表示容量为j时能放的最多东西。从1到n个物品一个一个的放和比较。

#include<cstdio>
#include<cstring>
using namespace std;
int main()
{
    int w[3500],d[3500],n,m,i,j,dp[13500];
    while (~scanf("%d%d",&n,&m))
    {
        memset(dp,0,sizeof(dp));
        for (i=1;i<=n;i++) scanf("%d%d",&w[i],&d[i]);
        for (i=1;i<=n;i++)
        {
            for (j=m;j>=w[i];j--)
            if (dp[j]<dp[j-w[i]]+d[i])
            dp[j]=dp[j-w[i]]+d[i];
        }
        printf("%d
",dp[m]);
    }
}