【HDU 2602】Bone Collector（01背包）

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

题目大意

背包问题，给你两个数N，V分别表示骨头的个数和背包的容量，接下来一行表示每个骨头的价值，在接下来的一行表示每个骨头的重量。求最大能被价值为多少的骨头。

思路

01背包裸体

代码

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn=1000+5;int w[maxn],v[maxn],dp[maxn],t,n,cost;int main(){    scanf("%d",&t);    while(t--)    {        memset(dp,0,sizeof(dp));        scanf("%d %d",&n,&cost);        for(int i=1;i<=n;i++)        {            scanf("%d",&v[i]);        }        for(int i=1;i<=n;i++)        {            scanf("%d",&w[i]);        }        for(int i=1;i<=n;i++)        {            for(int j=cost;j>=w[i];j--)            {                dp[j]=max(dp[j],dp[j-w[i]]+v[i]);            }        }        printf("%d\n",dp[cost]);    }    return 0;}