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2013
11-13

POJ 3627 Bookshelf [解题报告] Java

Bookshelf

问题描述 :

Farmer John recently bought a bookshelf for cow library, but the shelf is getting filled up quite quickly, and now the only available space is at the top.

Each of the N cows (1 ≤ N ≤ 20,000) has some height of Hi (1 ≤ Hi ≤ 10,000) and a total height summed across all N cows of S. The bookshelf has a height of B (1 ≤ BS < 2,000,000,007).

To reach the top of the bookshelf taller than the tallest cow, 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. Since more cows than necessary in the stack can be dangerous, your job is to find the set of cows that produces a stack of the smallest number of cows possible such that the stack can reach the bookshelf.

输入:

* Line 1: Two space-separated integers: N and B
* Lines 2..N+1: Line i+1 contains a single integer: Hi

输出:

* Line 1: A single integer representing the size of the smallest set of cows that can reach the bookshelf.

样例输入:

6 40
6
18
11
13
19
11

样例输出:

3

解题代码:

//* @author 
import java.io.*; 
import java.util.*; 
import java.math.*; 
public class Main 
{ 
    static int[] bookHeight; 
    static int n; 
    static BigInteger b; 
    public static void main(String[] args) throws Exception 
    { 
        readFile(); 
    }    
    public static void readFile() throws Exception 
    { 
        BufferedReader br = new BufferedReader( 
            new InputStreamReader(System.in)); 
        StringTokenizer st = new StringTokenizer( 
            br.readLine()," "); 
        n = Integer.valueOf(st.nextToken()); 
        b = new BigInteger(st.nextToken()); 
        int iCount = 0; 
        bookHeight = new int[n]; 
        while(iCount< n) 
        { 
            bookHeight[iCount++] = Integer.valueOf( 
                br.readLine());  
        } 
        process(); 
    } 
    public static void defbug() 
    { 
        for(int i=0; i< bookHeight.length; i++) 
            System.out.println(bookHeight[i]+" "); 
        System.out.println(); 
    } 
    public static void process() 
    { 
        Arrays.sort(bookHeight); 
        BigInteger sum = BigInteger.valueOf(0L); 
        BigInteger temp; 
        int iCount = 0; 
        for(int i=bookHeight.length-1; i>=0; i--) 
        { 
            sum = sum.add(BigInteger.valueOf( 
                    (long)(bookHeight[i]))); 
            temp = sum.max(b); 
            iCount++; 
            if(temp.equals(sum)) 
                break;   
        } 
        System.out.print(iCount); 
    } 
}

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  2. #include <cstdio>

    int main() {
    //answer must be odd
    int n, u, d;
    while(scanf("%d%d%d",&n,&u,&d)==3 && n>0) {
    if(n<=u) { puts("1"); continue; }
    n-=u; u-=d; n+=u-1; n/=u;
    n<<=1, ++n;
    printf("%dn",n);
    }
    return 0;
    }

  3. 可以根据二叉排序树的定义进行严格的排序树创建和后序遍历操作。如果形成的排序树相同,其树的前、中、后序遍历是相同的,但在此处不能使用中序遍历,因为,中序遍历的结果就是排序的结果。经在九度测试,运行时间90ms,比楼主的要快。