hdu 2602 Bone Collector（01背包）

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 42179    Accepted Submission(s): 17543

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   |   We have carefully selected several similar problems for you:  2159 2955 1171 2844 1087 

 

很久之前就写过这道题，很久之前也接触过背包，写过很多01背包的题，但是就是记不住！貌似对所有的dp我都不能灵活的运用，因此感到非常懊恼。于是我决定把所以的背包题重新写一次。

最简单的背包题，模板01背包。

 

题意：第一行输入几组数据，第二行第一个数字代表物体个数，第二个数代表总体积。需要注意的是，第三排输入的是物品的价值，第四排的物品的体积。在不可以拆分物体的前提下，已知背包的总体积，最大能获取的价值是多少。

 

附上代码：

 

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int max(int a,int b)
{
    return a>b?a:b;
}
int main()
{
    int T,n,m,i,j;
    int a[1005],b[1005];
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        for(i=0; i<n; i++)    //输入价值
            scanf("%d",&a[i]);
        for(i=0; i<n; i++)   //输入体积
            scanf("%d",&b[i]);
        int dp[1005];  //dp数组始终记录当前体积的最大价值
        memset(dp,0,sizeof(dp));  //dp全部初始化为0
        for(i=0; i<n; i++)   //从第一个开始循环
            for(j=m; j>=b[i]; j--)
                dp[j]=max(dp[j],dp[j-b[i]]+a[i]);  //比较放入i物体后的价值与不放之前的价值，记录大的值
        printf("%d
",dp[m]);   //输入总体积的最大价值
    }
    return 0;
}