Day9 - C - Bookshelf 2 POJ - 3628

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 H_i (1 ≤ H_i ≤ 1,000,000 - these are very tall cows). The bookshelf has a height of B (1 ≤ B ≤ S, where S is 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: H_i

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

思路：读题读的头大，基础dp问题，01背包，dp[i][j]表示前i只牛，牛的高度为j，其能达到书架的最高高度，转移方程为选/不选第i只牛，dp[i][j]=max(dp[i-1][j],dp[i-1][j-h[i]]+h[i])

const int maxm = 2e7+;

int dp[maxm], sum, buf[];

int main() {
    ios::sync_with_stdio(false), cin.tie();
    int N, B;
    cin >> N >> B;
    for(int i = ; i <= N; ++i) {
        cin >> buf[i];
        sum += buf[i];
    }
    for(int i = ; i <= N; ++i) {
        for(int j = sum; j >= buf[i]; --j)
            dp[j] = max(dp[j], dp[j-buf[i]]+buf[i]);
    }
    for(int i = ; i <= sum; ++i)
        if(dp[i] >= B) {
            cout << dp[i] - B << endl;
            break;
        }

    return ;
}