HDU2602-Bone Collector（01背包）

骨收集者的最大价值

最新推荐文章于 2022-04-03 11:21:11 发布

原创最新推荐文章于 2022-04-03 11:21:11 发布 · 133 阅读

1 ·

CC 4.0 BY-SA版权

一般DP/背包专栏收录该内容

5 篇文章

订阅专栏

本文介绍了一道经典的01背包问题，通过二维DP和一维DP两种方法解决骨收集者如何在有限背包容量下获得最大价值的问题。文章提供了完整的代码实现，帮助读者理解状态转移方程。

题目链接

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

5 10

1 2 3 4 5

5 4 3 2 1

Sample Output

裸的01背包，有二维dp，和一维dp两种方法。

状态转移方程： f[i][j] = max（ f[i-1][j]，f[i-1][ j-w[i] ]+v[i]）

//二维dp
#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;
const int Max = 1010;
int v[Max],w[Max],f[Max][Max];
int t,x,y,m,n;

int main()
{
    scanf("%d",&t);
    while(t--)
    {
        memset(v,0,sizeof(v));
        memset(w,0,sizeof(w));
        memset(f,0,sizeof(f));
        scanf("%d %d",&m,&n);
        for(int i=1;i<=m;i++) scanf("%d",&v[i]);
        for(int i=1;i<=m;i++) scanf("%d",&w[i]);
        for(int i=1;i<=m;i++){
            for(int j=0;j<=n;j++){ //注意物品重量可能为0
                if(j>=w[i]) f[i][j] = max(f[i-1][j],f[i-1][j-w[i]]+v[i]);
                else f[i][j] = f[i-1][j];
            }
        }
        printf("%d\n",f[m][n]);
    }
    return 0;
}

//一维dp
#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;
const int Max = 1010;
int v[Max],w[Max],f[Max];
int t,x,y,m,n;

int main()
{
    scanf("%d",&t);
    while(t--)
    {
        memset(v,0,sizeof(v));
        memset(w,0,sizeof(w));
        memset(f,0,sizeof(f));
        scanf("%d %d",&m,&n);
        for(int i=1;i<=m;i++) scanf("%d",&v[i]);
        for(int i=1;i<=m;i++) scanf("%d",&w[i]);
        for(int i=1;i<=m;i++){
            for(int j=n;j>=0;j--){
                if(j>=w[i]) f[j] = max(f[j],f[j-w[i]]+v[i]);
            }
        }
        printf("%d\n",f[n]);
    }
    return 0;
}