bookshelf 2 (01 가방 입문 또는 dfs)

Bookshelf 2
Time Limit: 1000MS
 
Memory Limit: 65536K
Total Submissions: 4064
 
Accepted: 1857
Description
Farmer John recently bought another bookshelf for the cow library, but the shelf is getting filled up quite quickly, and now the only available space is at the top.
FJ has N cows (1 ≤ N ≤ 20) each with some height of Hi (1 ≤ Hi ≤ 1,000,000 - these are very tall cows). The bookshelf has a height of B (1 ≤ B ≤ S, where Sis the sum of the heights of all cows).
To reach the top of the bookshelf, one or more of the cows can stand on top of each other in a stack, so that their total height is the sum of each of their individual heights. This total height must be no less than the height of the bookshelf in order for the cows to reach the top.
Since a taller stack of cows than necessary can be dangerous, your job is to find the set of cows that produces a stack of the smallest height possible such that the stack can reach the bookshelf. Your program should print the minimal 'excess' height between the optimal stack of cows and the bookshelf.
Input
* Line 1: Two space-separated integers: N and B * Lines 2..N+1: Line i+1 contains a single integer: Hi
Output
* Line 1: A single integer representing the (non-negative) difference between the total height of the optimal set of cows and the height of the shelf.
Sample Input

5 16
3
1
3
5
6

Sample Output

1

Source
USACO 2007 December Bronze
제목:http://poj.org/problem?id=3628
분석: 이 문제는 01 가방으로 풀면 정말 억지스럽다. dfs처럼 보이지만 DP의 코드는 매우 짧다
마찬가지로 f[i]를 설정하면 i의 높이가 도달할 수 있는지를 나타낸다. f[i]=1(f[j]=1, j+h[k]=i), 이런 가방은 두 번째 종류에 속한다. 즉, 가방의 용량을 엄격하게 제한하고 채워야 한다. 초기값을 부여할 때 f[0]만 도달할 수 있고 다른 것은 모두 도달할 수 없다. 이동할 때 f[j]가 도달할 수 있는지 판단해야 한다...
코드:

#include<cstdio>
using namespace std;
const int mm=20000000;
bool f[mm];
int h[22];
int i,j,k,n,b,m;
int main()
{
    while(scanf("%d%d",&n,&b)!=-1)
    {
        for(m=i=0;i<n;++i)scanf("%d",&h[i]),m+=h[i];
        for(i=0;i<=m;++i)f[i]=0;
        f[0]=1;
        for(i=0;i<n;++i)
            for(j=m-h[i];j>=0;--j)
                if(f[j])f[j+h[i]]=1;
        for(i=b;i<=m;++i)
            if(f[i])break;
        printf("%d
",i-b);
    }
    return 0;
}