Bone Collector（01背包问题）

最新推荐文章于 2020-02-12 18:53:54 发布

爱上键盘的小哥哥

最新推荐文章于 2020-02-12 18:53:54 发布

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分类专栏：背包问题DP

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本文介绍了一个经典的01背包问题——骨值最大化问题。该问题要求在限定的背包体积内，选择若干骨头放入背包中，使得骨头总价达到最大值。文章提供了完整的C++代码实现。

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 2 ³¹).

Sample Input

Sample Output

题意：

有一个人喜欢收集各种骨头，要在往一个固定体积的背包里放骨头，每块骨头的价值和体积不同；

求背包里骨头价值的总和的最大值

思路：裸的01背包问题

代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define MAX 1005
using namespace std;
long bag[MAX];
long val[MAX];
long w[MAX];
int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        memset(bag,0,sizeof(bag));
        memset(val,0,sizeof(val));
        memset(w,0,sizeof(w));
        int n,m;
        scanf("%d%d",&n,&m);
        int i;
        for(i=1;i<=n;i++)//记录每块骨头的价值
            scanf("%ld",&val[i]);
        for(i=1;i<=n;i++)//记录每块骨头的体积
            scanf("%ld",&w[i]);
        int j;
        for(i=1;i<=n;i++){  //背包算法核心
            for(j=m;j>=0;j--){
                if(w[i]<=j){
                    bag[j]=max(bag[j],bag[j-w[i]]+val[i]);
                }
            }
        }
        printf("%ld\n",bag[m]);
    }
return 0;
}