HDU 2602

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2602

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 ? 

InputThe 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. OutputOne 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

题解：

01背包模板题。

AC代码：

二维数组写法：

#include<cstdio>
#include<algorithm>
using namespace std;

unsigned long max_weight[1003][1003];//max_weight[i][j]:前i个骨头中，容量为j的背包里能放的下的最大重量 
struct type{
    int v;
    int w;
}bone[1003];

int main()
{
    int n,V;
    int t;scanf("%d",&t);
    while(t--){
        scanf("%d%d",&n,&V);
        for(int i=1;i<=n;i++) scanf("%d",&bone[i].w);
        for(int i=1;i<=n;i++) scanf("%d",&bone[i].v);
        
        for(int j=0;j<=V;j++){
            if(bone[1].v <= j) max_weight[1][j]=bone[1].w;
            else max_weight[1][j]=0;
        }
        
        for(int i=2;i<=n;i++){
            for(int j=0;j<=V;j++){
                if(j<bone[i].v) max_weight[i][j]=max_weight[i-1][j];
                else max_weight[i][j]=max( max_weight[i-1][j] , max_weight[i-1][ (j-bone[i].v) ] + bone[i].w );
            }
        }
        
        printf("%d
",max_weight[n][V]);
    }
}

减小空间复杂度，滚动一维数组写法：

#include<cstdio>
#include<algorithm>
using namespace std;

unsigned long max_weight[1003];
struct type{
    int v;
    int w;
}bone[1003];

int main()
{
    int n,V;
    int t;scanf("%d",&t);
    while(t--){
        scanf("%d%d",&n,&V);
        for(int i=1;i<=n;i++) scanf("%d",&bone[i].w);
        for(int i=1;i<=n;i++) scanf("%d",&bone[i].v);
        
        for(int j=0;j<=V;j++){
            if(bone[1].v <= j) max_weight[j]=bone[1].w;
            else max_weight[j]=0;
        }
        
        for(int i=2;i<=n;i++){
            for(int j=V;j>=0;j--){
                if(j<bone[i].v) max_weight[j]=max_weight[j];
                else max_weight[j]=max( max_weight[j] , max_weight[ (j-bone[i].v) ] + bone[i].w );
            }
        }
        
        printf("%d
",max_weight[V]);
    }
}

空间复杂度的对比：