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

Sample Output

题解：

01背包模板题。

AC代码：

二维数组写法：

 1 #include<cstdio>
 2 #include<algorithm>
 3 using namespace std;
 4 
 5 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 
 6 struct type{
 7     int v;
 8     int w;
 9 }bone[1003];
10 
11 int main()
12 {
13     int n,V;
14     int t;scanf("%d",&t);
15     while(t--){
16         scanf("%d%d",&n,&V);
17         for(int i=1;i<=n;i++) scanf("%d",&bone[i].w);
18         for(int i=1;i<=n;i++) scanf("%d",&bone[i].v);
19         
20         for(int j=0;j<=V;j++){
21             if(bone[1].v <= j) max_weight[1][j]=bone[1].w;
22             else max_weight[1][j]=0;
23         }
24         
25         for(int i=2;i<=n;i++){
26             for(int j=0;j<=V;j++){
27                 if(j<bone[i].v) max_weight[i][j]=max_weight[i-1][j];
28                 else max_weight[i][j]=max( max_weight[i-1][j] , max_weight[i-1][ (j-bone[i].v) ] + bone[i].w );
29             }
30         }
31         
32         printf("%d
",max_weight[n][V]);
33     }
34 }

减小空间复杂度，滚动一维数组写法：

 1 #include<cstdio>
 2 #include<algorithm>
 3 using namespace std;
 4 
 5 unsigned long max_weight[1003];
 6 struct type{
 7     int v;
 8     int w;
 9 }bone[1003];
10 
11 int main()
12 {
13     int n,V;
14     int t;scanf("%d",&t);
15     while(t--){
16         scanf("%d%d",&n,&V);
17         for(int i=1;i<=n;i++) scanf("%d",&bone[i].w);
18         for(int i=1;i<=n;i++) scanf("%d",&bone[i].v);
19         
20         for(int j=0;j<=V;j++){
21             if(bone[1].v <= j) max_weight[j]=bone[1].w;
22             else max_weight[j]=0;
23         }
24         
25         for(int i=2;i<=n;i++){
26             for(int j=V;j>=0;j--){
27                 if(j<bone[i].v) max_weight[j]=max_weight[j];
28                 else max_weight[j]=max( max_weight[j] , max_weight[ (j-bone[i].v) ] + bone[i].w );
29             }
30         }
31         
32         printf("%d
",max_weight[V]);
33     }
34 }

空间复杂度的对比：