HDU 2602 - Bone Collector - [01背包模板题]

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[][];//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[];

 int main()
 {
     int n,V;
     int t;scanf("%d",&t);
     while(t--){
         scanf("%d%d",&n,&V);
         for(int i=;i<=n;i++) scanf("%d",&bone[i].w);
         for(int i=;i<=n;i++) scanf("%d",&bone[i].v);

         for(int j=;j<=V;j++){
             if(bone[].v <= j) max_weight[][j]=bone[].w;
             else max_weight[][j]=;
         }

         for(int i=;i<=n;i++){
             for(int j=;j<=V;j++){
                 if(j<bone[i].v) max_weight[i][j]=max_weight[i-][j];
                 else max_weight[i][j]=max( max_weight[i-][j] , max_weight[i-][ (j-bone[i].v) ] + bone[i].w );
             }
         }

         printf("%d\n",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[];
 struct type{
     int v;
     int w;
 }bone[];

 int main()
 {
     int n,V;
     int t;scanf("%d",&t);
     while(t--){
         scanf("%d%d",&n,&V);
         for(int i=;i<=n;i++) scanf("%d",&bone[i].w);
         for(int i=;i<=n;i++) scanf("%d",&bone[i].v);

         for(int j=;j<=V;j++){
             if(bone[].v <= j) max_weight[j]=bone[].w;
             else max_weight[j]=;
         }

         for(int i=;i<=n;i++){
             for(int j=V;j>=;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\n",max_weight[V]);
     }
 }

空间复杂度的对比：