hdu2602 01背包Bone Collector

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 40316 Accepted Submission(s): 16748



[align=left]Problem Description[/align]
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 ?

[align=left]Input[/align]
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.

[align=left]Output[/align]
One integer per line representing the maximum of the total value (this number will be less than 231).

[align=left]Sample Input[/align]

1
5 10
1 2 3 4 5
5 4 3 2 1

[align=left]Sample Output[/align]

14本人还没有学c++

[code]

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct stu {
int val;
int cos;
}boy[10010];
//stu ;
//int bag[10010];
int nexty[10010];//如果把所有nexty都改为next提交时出现编译错误，应该是因为next数组在c++里面已经有定义（本人还没有学c++）。
int main()
{
int n;
int a,b;
scanf("%d",&n);
while(n--)
{
scanf("%d%d",&a,&b);
memset(boy,0,sizeof(boy));
//	memset(bag,0,sizeof(bag));
memset(nexty,0,sizeof(nexty));
for(int i=0;i<a;i++)
scanf("%d",&boy[i].val);
for(int j=0;j<a;j++)
scanf("%d",&boy[j].cos);
for(int i=0;i<a;i++)
{
for(int j=b;j>=boy[i].cos;j--)
//	bag[j]=max(bag[j],bag[j-boy[i].cos]+boy[i].val);
nexty[j]=max(nexty[j],nexty[j-boy[i].cos]+boy[i].val);
}
//printf("%d\n",bag[b]);
printf("%d\n",nexty[b]);
}
return 0;
}

[/code]