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：

HDU 2602 - Bone Collector - [01背包模板题]的相关教程结束。