Bookshelf 2 POJ - 3628（01背包||DFS）

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

Sample Output

代码：搜索

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int digit[25];
int book[25];
int gg[25];
typedef long long ll;
int n,b;
int inf=0x3f3f3f3f;
void DFS(int cnt,int sum)
{if(sum+gg[n]-gg[cnt-1]<b)return;if(sum>=b){inf=min(inf,sum);return;}DFS(cnt+1,sum);DFS(cnt+1,sum+digit[cnt]);return;
}
int main()
{   scanf("%d%d",&n,&b);for(int i=1;i<=n;i++){scanf("%d",&digit[i]);gg[i]=gg[i-1]+digit[i];}   DFS(1,0);printf("%d\n",inf-b);return 0;
}

代码：DP

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int digit[25];
typedef long long ll;
int n,b;
int sum=0;
int inf=0x3f3f3f3f;
int dp[1000000*20+5];
int main()
{   scanf("%d%d",&n,&b);for(int i=1;i<=n;i++){scanf("%d",&digit[i]);sum=sum+digit[i];} for(int i=1;i<=n;i++){for(int j=sum;j>=digit[i];j--){dp[j]=max(dp[j],dp[j-digit[i]]+digit[i]); }}for(int i=1;i<=sum;i++){if(dp[i]>=b){cout<<dp[i]-b<<endl;break;}}return 0;
}

Bookshelf 2 POJ - 3628（01背包||DFS）相关推荐

【 POJ - 3628 】Bookshelf 2（dfs 或 dp，0-1背包）
题干: Farmer John recently bought another bookshelf for the cow library, but the shelf is getting fill ...
POJ 3628 Bookshelf 2
POJ 3628 Bookshelf 2:http://poj.org/problem?id=3628 题意:有个书架,高度为B,现在FJ有N个奶牛,每个奶牛有个高度hi,现在将奶牛堆起来,使得堆起来 ...
poj 2063 Investment（01背包变形）
http://poj.org/gotoproblem?pid=2063 (1)上限 m 一直上升的 n 次01背包问题,比一般的01背包多了一重循环: (2)本题出现了各种错误:1)刚开始,没注意 m ...
bzoj2287【POJ Challenge】消失之物缺一01背包
bzoj2287[POJ Challenge]消失之物缺一01背包链接 bzoj 思路分治solve(l,r,arr)表示缺少物品\([l,r]\)的dp数组arr. 然后solve(l,mid ...
【POJ - 3211】Washing Clothes （dp，0-1背包中点问题）
题干: Dearboy was so busy recently that now he has piles of clothes to wash. Luckily, he has a beautif ...
【洛谷 - U43391】不是0-1背包的暴力AC（思维，二分，可转化为二元组问题，复习暴力dfs总结）
题干: https://www.luogu.org/problemnew/show/U43391 自01背包问世之后,小A对此深感兴趣.一天,小A去远游,却发现他的背包不同于01背包. 小A的背包最多 ...
LeetCode 1774. 最接近目标价格的甜点成本（DFS / 01背包）
文章目录 1. 题目 2. 解题 1. 题目你打算做甜点,现在需要购买配料.目前共有 n 种冰激凌基料和 m 种配料可供选购.而制作甜点需要遵循以下几条规则: 必须选择一种冰激凌基料. 可以添加 ...
POJ3628:Bookshelf 2【01背包】
Description Farmer John recently bought another bookshelf for the cow library, but the shelf is gett ...
POJ 3624 Charm Bracelet 0-1背包
传送门:http://poj.org/problem?id=3624 题目大意:XXX去珠宝店,她需要N件首饰,能带的首饰总重量不超过M,要求不超过M的情况下,使首饰的魔力值(D)最大. 0-1背包入 ...
hdoj 1864 最大报销额【01背包】||【dfs】
最大报销额 Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Subm ...

Bookshelf 2 POJ - 3628（01背包||DFS）

Bookshelf 2

Bookshelf 2 POJ - 3628（01背包||DFS）相关推荐

最新文章

热门文章