hdu 2602 Bone Collector (01背包经典入门)

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2602

                                         Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 13839    Accepted Submission(s): 5454

Problem Description

Many years ago , in Teddy’s hometown there was a man who was called “Bone Collector”. This man like to collect varies of bones , such as dog’s , cow’s , also he went to the grave …
The bone collector had a big bag with a volume of V ,and along his trip of collecting there are a lot of bones , obviously , different bone has different value and different volume, now given the each bone’s value along his trip , can you calculate out the maximum of the total value the bone collector can get ?

 

Input

The first line contain a integer T , the number of cases.
Followed by T cases , each case three lines , the first line contain two integer N , V, (N <= 1000 , V <= 1000 )representing the number of bones and the volume of his bag. And the second line contain N integers representing the value of each bone. The third line contain N integers representing the volume of each bone.

 

Output

One integer per line representing the maximum of the total value (this number will be less than 2³¹).

 

Sample Input

1 5 10 1 2 3 4 5 5 4 3 2 1

 

Sample Output

14

 

Author

Teddy

 

Source

HDU 1st “Vegetable-Birds Cup” Programming Open Contest

 

Recommend

lcy

 

很基础的背包入门题目，如果看了代码还不会做，就自己去百度文档中下载《背包九讲》吧。很强大的，注意有个事不要积分就可以下载的，看清楚了啊！

给几个下载地址：http://wenku.baidu.com/view/16527702de80d4d8d15a4f06.html

                               http://wenku.baidu.com/view/16527702de80d4d8d15a4f06.html

//hdu 2602 AC 0ms 248k

#include<stdio.h>
#include<string.h>
int main()
{
    int test;
    int n,v; 
    int i,j;
    int c[1005],w[1005],sum[1005];
    scanf("%d",&test);
    while(test--)
    {
        memset(sum,0,sizeof(sum));
        scanf("%d%d",&n,&v);
        for(i=0;i<n;i++)
            scanf("%d",&w[i]);//输入价值
        for(i=0;i<n;i++)
            scanf("%d",&c[i]);//输入对应的体积
        for(i=0;i<n;i++)
            for(j=v;j>=c[i];j--)
            {
                if(sum[j]<=(sum[j-c[i]]+w[i]))
                    sum[j]=sum[j-c[i]]+w[i];
            }
        printf("%d\n",sum[v]);
    }
    return 0;
}

下面再贴个另外的代码，千万别提交啊，不是这道题目的。

//注意下面的不要提交啊~~~ 
//不过貌似是另外一道题目的AC代码，具体是哪道题目我忘了。。。
//看了kb神的另外一篇完全背包博客后从新写的，如果改变测试数据输入的顺序就可如此AC，应该可以省下不少内存，虽然内存不要钱，但是能省则省了哈！O(∩_∩)O哈哈哈~大神果然强大 
#include<stdio.h>
#include<string.h>
int main()
{
    int test;
    int n,v;
    int i,j;
    int c,w,sum[1005];
    scanf("%d",&test);
    while(test--)
    {
        memset(sum,0,sizeof(sum));
        scanf("%d%d",&n,&v);
        while(n--)
        {
            scanf("%d%d",&w,&c);
            for(i=v;i>=c;i--)
                if(sum[i]<=(sum[i-c]+w))
                   sum[i]=sum[i-c]+w;
        }
        printf("%d\n",sum[v]);
    }
    return 0;
}