HDUOJ--Bone Collector
Bone Collector
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 21463 Accepted Submission(s): 8633
Problem Description
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 231).
Sample Input
1 5 10 1 2 3 4 5 5 4 3 2 1
Sample Output
14
Author
Teddy
Source
http://acm.hdu.edu.cn/showproblem.php?pid=2602
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代码：----
1 #include<stdio.h>
2 #include<string.h>
3 #define maxn 1005
4 int dp[maxn],arr[maxn][2];
5 int max(int a,int b)
6 {
7 return a>b?a:b;
8 }
9
10 void zeroonepack(int cost ,int value,int v)
11 {
12 for(int i=v;i>=cost;i--)
13 dp[i]=max(dp[i],dp[i-cost]+value);
14
15 }
16 int main()
17 {
18 int t,n,v ,i;
19 scanf("%d",&t);
20 while(t--)
21 {
22 scanf("%d%d",&n,&v);
23 memset(dp,0,sizeof dp);
24 for(i=0;i<n;i++)
25 {
26 scanf("%d",arr[i]+0);
27 }
28 for(i=0;i<n;i++)
29 {
30 scanf("%d",arr[i]+1);
31 }
32 for(i=0;i<n;i++)
33 zeroonepack(arr[i][1],arr[i][0],v);
34 printf("%d\n",dp[v]);
35 }
36 return 0;
37 }优化后代码：
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
struct st
{
int a;
int b;
};
typedef struct st sta;
int main()
{
int test,n,v,i,j;
scanf("%d",&test);
while(test--)
{
scanf("%d%d",&n,&v);
int *dp =(int *)malloc(sizeof(int)*(v+1));
sta *stu =(sta *)malloc(sizeof(sta)*(n+1));
for(i=0;i<n;i++)
scanf("%d",&stu[i].a);
for(i=0;i<n;i++)
scanf("%d",&stu[i].b);
for(i=0;i<=v;i++)
dp[i]=0;
for(i=0;i<n;i++)
{
for(j=v ; j>=stu[i].b ; j--)
{
if(dp[j]<dp[j-stu[i].b]+stu[i].a)
dp[j]=dp[j-stu[i].b]+stu[i].a;
}
}
printf("%d\n",dp[v]);
free(dp);
free(stu);
}
return 0;
}