HDU 2602 Bone Collector(01背包裸题)

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 thereare 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

思路：

动规01 背包模板题，真的是裸的不能再裸了。

代码：

#include<stdio.h>
#include<iostream>
#include<queue>
#include<cstring>
#include<cmath>

using namespace std;

#define MAXN 1005

int N,V;
int vol[MAXN],val[MAXN];
int dp[MAXN];

int main()
{
	int T;
	scanf("%d",&T);
	while(T--)
	{
		memset(dp,0,sizeof(dp));
		scanf("%d %d",&N,&V);
		for(int i=0 ; i<N ; i++)
		{
			scanf("%d",&val[i]);
		}
		for(int i=0 ; i<N ; i++)
		{
			scanf("%d",&vol[i]);
		}
		for(int i=0 ; i<N ; i++)
		{
			for(int j=V ; j>=vol[i] ; j--)
			{
				dp[j] = max(dp[j],dp[j-vol[i]]+val[i]);
			}
		}
		printf("%d
",dp[V]);
	}
	return 0;
}