hdu2602（01背包）

HDU_2602 Bone Collector

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

Total Submission(s): 49986    Accepted Submission(s): 20965

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 2 ³¹).

 

Sample Input

  
  

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

 

Sample Output

  
  

   
   14
  
  

套0-1背包公式，直接写。

#include <bits/stdc++.h>
using namespace std;
const int maxn=1010;
int ans[maxn];
int value[maxn],volum[maxn];
int main(int argc,char *argv[])
{
//	freopen("data.in.txt","r",stdin);
	int ncase;
	cin>>ncase;
	while(ncase--){
		int n,v;
		cin>>n>>v;
		
		for(int i=1;i<=n;i++)
			cin>>value[i];
		for(int i=1;i<=n;i++)
			cin>>volum[i];
			
		memset(ans,0,sizeof(ans));
		for(int i=1;i<=n;i++)
			for(int j=v;j>=0;j--){
				if(j>=volum[i])
				ans[j]=max(ans[j],ans[j-volum[i]]+value[i]);
			}
		cout<<ans[v]<<endl;
	} 
	return 0;
}