hdu- 2602 Bone Collector

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).

这道题目是我做的第一道背包问题，以前没注意过，以后会多学这方面的东西。
核心代码：
 dp[j]=max(dp[j],dp[j-c[i]]+w[i]);
所有代码：
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int dp[1200];
int w[1200];
int c[1200];
int main(){
	int T;
	int i,j,k,t;
	int n,v;
	scanf("%d",&T);
	while(T--){
		scanf("%d%d",&n,&v);
		for(i=0;i<n;i++)
		  scanf("%d",&w[i]);
		for(i=0;i<n;i++)
		  scanf("%d",&c[i]);
		memset(dp,0,sizeof(dp));
		for(i=0;i<n;i++){
			for(j=v;j>=c[i];j--)
			  dp[j]=max(dp[j],dp[j-c[i]]+w[i]);
		} 
		printf("%d\n",dp[v]);
	}
	return 0;
}