zoukankan      html  css  js  c++  java
  • POJ 3628 Bookshelf 2

    Bookshelf 2

    http://acm.hdu.edu.cn/webcontest/contest_showproblem.php?pid=1002&ojid=1&cid=2967&hide=0

    Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)
    Total Submission(s) : 76   Accepted Submission(s) : 37
    Problem 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 Hi (1 ≤ Hi ≤ 1,000,000 - these are very tall cows). The bookshelf has a height of B (1 ≤ BS, 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: Hi

     
    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
     
    Source
    PKU
     
     
    #include<iostream>
    #include<cstdio>
    #include<cstring>
    
    using namespace std;
    
    const int INF=0x3f3f3f3f;
    
    int n,b;
    int cow[30];
    int dp[20000010];
    
    int main(){
    
        //freopen("input.txt","r",stdin);
    
        while(~scanf("%d%d",&n,&b)){
            int sum=0;
            for(int i=0;i<n;i++){
                scanf("%d",&cow[i]);
                sum+=cow[i];
            }
            for(int i=0;i<=sum;i++) //若使用memset则会导致内存超出限制
                dp[i]=0;
            for(int i=0;i<n;i++)
                for(int j=sum;j>=cow[i];j--)
                    if(dp[j]<dp[j-cow[i]]+cow[i])
                        dp[j]=dp[j-cow[i]]+cow[i];
    
    
            int ans=INF;
            for(int i=b;i<=sum;i++)
                if(dp[i]>=b){
                    ans=dp[i]-b;
                    break;
                }
            printf("%d\n",ans);
        }
        return 0;
    }
  • 相关阅读:
    类之OCP(Open Closed Principle):开闭原则
    OO书籍 zz
    ObjectOriented Design Heuristics (zz)
    CSS3 filter(滤镜) 网站整体变灰色调
    js和jquery设置css样式的几种方法
    20条书写CSS代码
    js 调用向html追加内容
    移动端检测微信浏览器返回,关闭,进入后台操作
    防止表单重复提交的4种方法
    CSS浮动标准修复top塌陷和清除浮动及IE兼容标准格式
  • 原文地址:https://www.cnblogs.com/jackge/p/2939656.html
Copyright © 2011-2022 走看看