poj3628Bookshelf 2

Bookshelf 2

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 8618 Accepted: 3924

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 1631356

Sample Output

1

Source

USACO 2007 December Bronze

在搜 背包的时候把这道 给搜出来了，仔细一看，用背包不简单，超时还超内存，题目给的是1000000，用循环去解决 太麻烦了，所以用了深搜.

附代码：

#include<string.h>#include<stdio.h>#include<algorithm>using namespace std;long ans[22][22],i,j,k,l,m,n,p;long smal=0x3f3f3f;long a[22];void dfs(int he,int ge){if(he>=n){if(he-n<smal)smal=he-n;}if(ge>m)return ;dfs(he+a[ge],ge+1);dfs(he,ge+1);}int main(){while(scanf("%ld%ld",&m,&n)!=EOF){for(i=1;i<=m;i++)scanf("%ld",&a[i]);dfs(0,1);printf("%ld\n",smal);}return 0;}