HDU 2602 Bone Collector【01背包入门题】

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 52370    Accepted Submission(s): 22079

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 2 ³¹).

 

Sample Input

  
  

   
   1
5 10
1 2 3 4 5
5 4 3 2 1
  
  

 

Sample Output

  
  

   
   14
  
  

 

Author

Teddy

 

Source

HDU 1st “Vegetable-Birds Cup” Programming Open Contest

 

Recommend

lcy

 

原题链接：http://acm.split.hdu.edu.cn/showproblem.php?pid=2602

01背包入门题：

转态转移方程：

f[i][v] = max ( f[i-1][v],f[i-1][v-c[i] ]+w[i] )

f[i][v]:前 i 件物品放入容量为v的背包获得最大价值。

c[i]: 第 i 件物品的体积。

w[i] :第 i 件物品的体积价值。

注意：输入数据的时候数组下标要从 1 开！

优化成一维的：

伪代码：

for i=1..N
   for v=V..0
        f[v]=max{f[v],f[v-c[i]]+w[i]};

f[v] : 体积为v的背包的最大价值。

参考博客：http://www.wutianqi.com/?p=539

http://www.cnblogs.com/jiangjun/archive/2012/05/08/2489590.html

二维AC代码：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn=1005;
int v[maxn];
int val[maxn];
int dp[maxn][maxn];
int main()
{
    int T,n,V;
    ios::sync_with_stdio(false);
    cin.tie(0);
    //freopen("data/2602.txt","r",stdin);
    cin>>T;
    while(T--)
    {
        cin>>n>>V;
        memset(dp,0,sizeof(dp));
        //memset(v,0,sizeof(v));
        //memset(val,0,sizeof(val));
        for(int i=1;i<=n;i++)
            cin>>val[i];
        for(int i=1;i<=n;i++)
            cin>>v[i];

        for(int i=1;i<=n;i++)
        {
            for(int j=0;j<=V;j++)
            {
                if(j>=v[i])
                    dp[i][j]=max(dp[i-1][j],dp[i-1][j-v[i]]+val[i]);
                else
                    dp[i][j]=dp[i-1][j];
            }
        }
        cout<<dp[n][V]<<endl;

    }
    return 0;
}

优化成一维AC代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,V,v[1005],val[1005],dp[1005];
        scanf("%d%d",&n,&V);
        int i,j,k;
        for(i=0; i<n; i++)
            scanf("%d",&val[i]);
        for(i=0; i<n; i++)
            scanf("%d",&v[i]);
        memset(dp,0,sizeof(dp));
        for(i=0; i<n; i++)
            for(j=V; j>=v[i]; j--)
                dp[j]=max(dp[j],dp[j-v[i]]+val[i]);
        printf("%d\n",dp[V]);
    }
    return 0;
}

尊重原创，转载请注明出处：http://blog.youkuaiyun.com/hurmishine