hdu 2602:Bone Collector(01背包)

http://acm.hdu.edu.cn/showproblem.php?pid=2602

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

题意分析：

有一个骨头收集者，有一个体积为V的口袋，每个骨头有自己的价值和体积，求最多能装入口袋骨头的价值是多少。

解题思路：

01背包，注意第一行是价值，第二行是体积。

#include <stdio.h>
#include <string.h>
#include <algorithm>
#define N 1020
using namespace std;
struct date{
	long long v;
	long long w;
}a[N];
long long dp[N];
int main()
{
	int t, n, m, i, j;
	scanf("%d", &t);
	while(t--)
	{
		memset(dp, 0, sizeof(dp));
		scanf("%d%d", &n, &m);	
		for(i=1; i<=n; i++)
			scanf("%lld", &a[i].v);
		for(i=1; i<=n; i++)
			scanf("%lld", &a[i].w);
		for(i=1; i<=n; i++)
			for(j=m; j>=a[i].w; j--)
				dp[j]=max(dp[j], dp[j-a[i].w]+a[i].v);
		printf("%lld
", dp[m]);
	}
	return 0;
}