POJ3628 Bookshelf 2(01背包+dfs)

Bookshelf 2

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8745		Accepted: 3974

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 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

题意：　求N个数中和比减16最小的。
5+6+3+3 - 16 = 1最小的是1；
01背包，背包的容量是所有的数累加起来的和sum，想想为什么呢，可以累加起来的和一定比给的那个数大，然后就从b-sum查找背包中的比b大的第一个数就ok了，
很棒的转换
dfs也能轻松搞定

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
using namespace std;
const int INF = 0x3f3f3f3f;
const int MAX = 2000000 + 10;
int dp[MAX],sum,a[MAX];

int main()
{
    int n,b;
    while(scanf("%d%d", &n,&b) != EOF)
    {
        for(int i = 1; i <= n; i++)
        {
            scanf("%d", &a[i]);
            sum += a[i];
        }
        memset(dp, 0, sizeof(dp));
        for(int i = 1; i <= n; i ++)
        {
            for(int j = sum; j >= a[i]; j--)
            {
                dp[j] = max(dp[j],dp[j - a[i]] + a[i]);
            }
        }
        for(int i = b; i <= sum; i++)
        {
            if(dp[i] >= b)
            {
                printf("%d
",dp[i] - b);
                break;
            }
        }

    }

    return 0;
}