POJ 3628 Bookshelf 2 （01背包）

Bookshelf 2

Time Limit: 1000MS

Memory Limit: 65536K

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

Sample Output

题意：有n头牛，已知每头牛的高度和书架高度，一头牛可以站在另一头牛身上，总高度是他们的高度之和。要求能够达到书架的顶端，即这些牛的总高度不低于书架高度，求满足条件的总高度的最小值，输出他们的差值。

解题思路：把书架的高度看做背包容量，将牛的高度看做物品的体积和价值，dp[i] 表示书架高度为i，牛的总高度不超过i的最大高度。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int dp[2000002], h[22];
int main()
{int n, m, i, j;while(~scanf("%d%d",&n,&m)){int sum = 0;memset(dp,0,sizeof(dp));for(i = 1; i <= n; i++){scanf("%d",&h[i]);sum += h[i];}for(i = 1; i <= n; i++)for(j = sum; j >= h[i]; j--)dp[j] = max(dp[j], dp[j - h[i]] + h[i]);int Min = sum;for(i = m; i <= sum; i++)if(dp[i] >= m && dp[i] - m < Min)Min = dp[i] - m;printf("%d\n",Min);}return 0;
}

也可以用搜索来写：

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
int h[22], ans, flag;
int n, m;
void dfs(int k, int s)
{if(s == m){ans = 0;return ;}if(s >= m){if(s - m < ans)ans = s - m;return ;}for(int i = k; i < n; i++){dfs(i+1,s+h[i]);}
}
int main()
{int i;while(cin >> n >> m){int sum = 0;flag = 0;for(i = 0; i < n; i++){cin >> h[i];sum += h[i];}if(sum == m){cout << "0" << endl;continue;}ans = sum;dfs(0,0);cout << ans << endl;}return 0;
}

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