Bone Collector 0-1背包问题

Posted 2021-01-31 -y-i-

tags:

篇首语：本文由小常识网(cha138.com)小编为大家整理，主要介绍了Bone Collector 0-1背包问题相关的知识，希望对你有一定的参考价值。

题目描述：

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 ?

输入：

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.

输出：

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

样例：

Sample Input

Sample Output

14
题目大意：
有一个人喜欢收集骨头（呃），现在有N根骨头，每根骨头对应着各自的体积和价值，这个人的背包容量为V，求能装的骨头的最大总价值。
代码：

#include<iostream>
#include<stdio.h>
#define max(a,b) a>b?a:b
struct bone{
    int wei,val;
};
int f[1002][1002];
bone temp[1005];
int value(int num,int v){
    for(int i=1;i<=num;++i){
        for(int j=0;j<=v;++j){
            if(j<temp[i-1].wei)
                f[i][j]=f[i-1][j];
            else
                f[i][j]=max(f[i-1][j-temp[i-1].wei]+temp[i-1].val,f[i-1][j]);
        }
    }
    return f[num][v];
}
using namespace std;
int main(){
    int n,num,v;
    scanf("%d",&n);
    for(int i=0;i<n;++i){
        scanf("%d%d",&num,&v);
        for(int j=0;j<num;++j) scanf("%d",&temp[j].val);
        for(int j=0;j<num;++j) scanf("%d",&temp[j].wei);
        printf("%d
",value(num,v));
    }
    return 0;
}