Bone Collector

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

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

Author

Teddy

Source

HDU 1st “Vegetable-Birds Cup” Programming Open Contest

代码：

//hdu 2602 Bone Collector
#include <stdio.h>
#include <string.h>
#include <algorithm>
#define max(a,b) (a>b?a:b)
using namespace std;
struct inin
{
	int val;//价值！
	int cos;//体积！
}boy[10001];
int bag[10001];//下表的大小个数个物品总价值！
int main()
{
	int T;
	int n,v;//n代表背包的个数，v代表背包的总体积！
	scanf("%d",&T);
	while(T--)
	{
		memset(boy,0,sizeof(boy));//保存某个物品的价值和体积！
		memset(bag,0,sizeof(bag));//保存i个物品的总价值！
		scanf("%d%d",&n,&v);//n代表背包的个数，v代表背包的总体积！
		for(int i=1;i<=n;i++)
		{
			scanf("%d",&boy[i].val);//价值
		}
		for(int j=1;j<=n;j++)
		{
			scanf("%d",&boy[j].cos);//体积
		}
		for(int i=1;i<=n;i++)
		{
			for(int j=v;j>=boy[i].cos;j--)//从后向前求背包能装的最大的价值
			{
				bag[j]=max(bag[j],bag[j-boy[i].cos]+boy[i].val);//看放与不放第i件物品他的价值哪个大，选价值较大的放进数组bag里面
			}//放的意思不是在目前放的最大的体积的基础上放与不放，而是在你目前的体积减去你将要放的那个背包的体积所得到的体积所能容纳的最大的价值的基础上才放的！
		}
		printf("%d\n",bag[v]);//最终遍历完所有的物品之后最后一个最大的价值就是所求！
	}
	return 0;
}

版权声明：本文为博主原创文章，未经博主允许不得转载。