poj 3624 Charm Bracelet(01 가방 입문문)

Charm Bracelet
Time Limit: 1000MS
 
Memory Limit: 65536K
Total Submissions: 10399
 
Accepted: 4667
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 * Lines 2..N+1: Line i+1 describes charm i with two space-separated integers: Wi and Di
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
USACO 2007 December Silver
제목:http://poj.org/problem?id=3624
분석: 이 문제는 01배낭의 입문 문제입니다. 단지 스크롤 수조에 사용해야 합니다. f[i]는 i의 무게를 사용하고 얻은 최대치를 나타냅니다.
즉, f[i]=max(f[j]+d[k]) {j+w[k]=i}
코드:

#include<cstdio>
#include<iostream>
using namespace std;
const int mm=13333;
int f[mm],w[mm],d[mm];
int i,j,k,n,m;
int main()
{
    while(scanf("%d%d",&n,&m)!=-1)
    {
        for(i=0;i<n;++i)scanf("%d%d",&w[i],&d[i]);
        for(i=0;i<m;++i)f[i]=0;
        for(i=0;i<n;++i)
            for(j=m;j>=w[i];--j)f[j]=max(f[j],f[j-w[i]]+d[i]);
        printf("%d
",f[m]);
    }
    return 0;
}