暑假集训第四周周三赛 E - Charm Bracelet 最大魅力

E - Charm Bracelet

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit  Status  Practice  POJ 3624

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

分析：

这是一道最简单的动规，有专门的模板，今天才知道，我用的叫做滚动数组，挺高大上一个名字

他有专门的状态转移方程：

if(f[j]<f[j-a[i]]+b[i])
        f[j]=f[j-a[i]]+b[i];

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#include<stdio.h>
#include<string.h>
int main()
{
    int n,m,a[3500],b[35000],f[13000],i,j,s;
    scanf("%d%d",&n,&m);
    memset(f,0,sizeof(f));
    for(i=0;i<n;i++)
        scanf("%d %d",&a[i],&b[i]);
    for(i=0;i<n;i++)
        for(j=m;j>=a[i];j--)
        if(f[j]<f[j-a[i]]+b[i])
        f[j]=f[j-a[i]]+b[i];
        printf("%d\n",f[m]);
}