Bone Collector（01背包）

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

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一维dp数组AC代码：

#include<cstdio>#include<iostream>#include<cstring>#include<cmath>#include<algorithm>using namespace std;const int MAX = 5010;int a[MAX];int dp[MAX];int n;struct node{    int value,V;}p[MAX];bool cmp(node a,node b){    return a.value>b.value;}int main(){    int t;    scanf("%d",&t);    while(t--)    {        int n,v;        memset(dp,0,sizeof(dp));        scanf("%d%d",&n,&v);        for(int i=0;i<n;i++)        {            scanf("%d",&p[i].value);        }        for(int i=0;i<n;i++)        {            scanf("%d",&p[i].V);        }        sort(p,p+n,cmp);        for(int i=0;i<n;i++)        {            for(int j=v;j>=p[i].V;j--)            {                dp[j]=max(dp[j],dp[j-p[i].V]+p[i].value);              }        }        printf("%d\n",dp[v]);    }    return 0;}

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二维dp数组AC代码：

#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>using namespace std;const int MAX = 2000+10;int dp[MAX][MAX];struct Node{    int V;    int value;}p[MAX];int main(){    int t;    scanf("%d",&t);    while(t--)    {        memset(dp,0,sizeof(dp));        int n,v;        scanf("%d%d",&n,&v);        for(int i=1;i<=n;i++)        {            scanf("%d",&p[i].value);        }        for(int i=1;i<=n;i++)        {            scanf("%d",&p[i].V);        }        for(int i=1;i<=n;i++)        {            for(int j=0;j<=v;j++)            {                if(j>=p[i].V)//如果还能继续装                dp[i][j]=max(dp[i-1][j],dp[i-1][j-p[i].V]+p[i].value);                else//剩余体积不足以继续装                dp[i][j]=dp[i-1][j];            }        }        printf("%d\n",dp[n][v]);    }    return 0;}