01背包问题解析-优快云博客

本文介绍了一个经典的01背包问题实例，通过一个骨骸收集者的故事情境来阐述如何利用编程解决这类问题。文章提供了完整的C++代码实现，并对算法的基本思路进行了说明。

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 ?
Problem <wbr>Q

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

题意：给出袋子的最大容量，和每个物品的价值和体积，求最多能装多少价值的物品；

解题思路：最基本的01背包问题；

感悟:今晚心情很不好。。。

代码：

#include

#include

#include

#define maxn 1010

using namespace std;

int main()

{

//freopen("in.txt","r",stdin);

    long long
t,n,v,date[maxn],V[maxn],dp[maxn];

scanf("%lld",&t);

while(t--)

    {

memset(dp,0,sizeof dp);

scanf("%lld%lld",&n,&v);

for(int i=1;i<=n;i++)

scanf("%lld",&date[i]);

for(int i=1;i<=n;i++)

scanf("%lld",&V[i]);

for(int i=1;i<=n;i++)

for(int j=v;j>=V[i];j--)

{

dp[j]=max(dp[j],dp[j-V[i]]+date[i]);

}

printf("%d\n",dp[v]);

    }

}