ACM--steps--3.3.1--Bone Collector

Bone Collector

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 486 Accepted Submission(s): 239

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

15 101 2 3 4 55 4 3 2 1

 

Sample Output

14

 

Author

Teddy

#include<iostream>#include<cstring>using namespace std;struct ws{    int val,wei;}dyx[1005];int wyx[1005][1005];//背包问题的dp方程。int main(){    int T;    cin>>T;    while(T--)    {        //该状态转移方程为；        //wyx[i][v]=max(dyx[i-1][v],dyx[i-1][v-i.wei]+i.val;        //意义为：在当前背包容量为v的情况下，第i个物品到底放不放。        memset(wyx,0,sizeof(wyx));        int n,v;//n表示骨头的数目，v表示背包的总容量。        cin>>n>>v;        for(int i=1;i<=n;i++)        cin>>dyx[i].val;        for(int i=1;i<=n;i++)        cin>>dyx[i].wei;        for(int i=1;i<=n;i++)        {            for(int j=0;j<=v;j++)//从背包从0的时候计算。            {                if(j-dyx[i].wei>=0&&wyx[i-1][j]<wyx[i-1][j-dyx[i].wei]+dyx[i].val)                //可以放进背包的话。                wyx[i][j]=wyx[i-1][j-dyx[i].wei]+dyx[i].val;                else                //放不进背包                wyx[i][j]=wyx[i-1][j];            }        }        cout<<wyx[n][v]<<endl;    }    return 0;}

一维数组的解法。

#include<iostream>#include<algorithm>#include<cstring>using namespace std;struct zr{    int wei,val;}wyx[1010];int dyx[1010];int main(){    int T;    cin>>T;    while(T--)    {        memset(dyx,0,sizeof(dyx));        int n,v;        cin>>n>>v;//N表示骨头的数目，V表示背包的容量。        for(int i=1;i<=n;i++)        cin>>wyx[i].val;        for(int i=1;i<=n;i++)        cin>>wyx[i].wei;        for(int i=1;i<=n;i++)        {            //在每次主循环中我们以v=V..0的顺序推dyx[v]，这样才能保证推dyx[v]时dyx[v-c[i]]保存的是状态dyx[i-1][v-c[i]]的值            for(int j=v;j>=wyx[i].wei;j--)            {                if(dyx[j]<dyx[j-wyx[i].wei]+wyx[i].val)                dyx[j]=dyx[j-wyx[i].wei]+wyx[i].val;            }        }        cout<<dyx[v]<<endl;    }    return 0;}