HDU 2602 Bone Collector

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 ?

InputThe 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.OutputOne integer per line representing the maximum of the total value (this number will be less than 2 ³¹).Sample Input

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

Sample Output

14


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

题意描述：

第一行输入骨头数N和背包的体积V (N <= 1000 , V <= 1000 )

第二行再输入N个整型数表示对应骨头的体积

最后第三行输入N个整型数表示对应骨头的价值

    解题思路：
        属于背包问题中的01背包问题，是背包问题的基础，也是能否解决好背包问题的关键。

01背包问题，顾名思义，有N件物品，有对应的体积和价值，问怎么放才能使体积为V的背包中所装的物品价值最大，那么对于每件物品有两个选择，不放入或者放入背包，对应状态为0或者1。

        解决问题的总体思路是运用动态规划的思想，对于每一件物品，先判断能否装的下，不能装的下，当前前i件物品放入体积为V的背包，最大价值等于前i-1件物品放入体积为V的背包的价值；能放得下判断

放进去价值更大还是不放进去价值更大。下面的代码使用优化后的方式实现，优化就是使用将二位数组优化为一维数组，遍历时将体积逆置。

    代码实现：

#include<stdio.h>
#include<string.h>
struct B
{
    int w,v;
};
int main()
{
    int i,t,N,V,f[1010],j;
    struct B b[1010];
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&N,&V);//注意读题，哪个是体积，哪个是价值 
        for(i=0;i<N;i++)
            scanf("%d",&b[i].v);
        for(i=0;i<N;i++)
            scanf("%d",&b[i].w);
        memset(f,0,sizeof(f));//全部初始化，不能使用多少初始化多少 
        for(i=0;i<N;i++){
            for(j=V;j>=0;j--){
                if(j>=b[i].w) 
                f[j] = f[j] > f[j-b[i].w]+b[i].v ? f[j] : f[j-b[i].w]+b[i].v;
            }
        }
        printf("%d
",f[V]);
    }
    return 0;
}

   易错分析：

1、注意读入数据时哪个是体积哪个是价值。

2、注意将f数组全部初始化，不然可能导致无法预计的错误。