HDU 2602 Bone Collector（01背包）

Bone Collector

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

Total Submission(s): 19446    Accepted Submission(s): 7706

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
   
   
   
   

 

Author

Teddy

 

Source

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

 

Recommend

lcy

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

//二维数组实现01背包
//用二维数组实现有个大坑  特大的坑  坑死了
#include<cstring>
#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
#define N 1005
int vol[N];
int val[N];
int dp[N][N];
int main ()
{
	int T;
	scanf("%d", &T);
	while(T--)
	{
		int n, v;
		scanf("%d%d", &n, &v);
		for(int i = 1; i <= n; i++)
			scanf("%d", &val[i]);
		for(int i = 1; i <= n; i++)
			scanf("%d", &vol[i]);
		memset(dp, 0, sizeof(dp));
		for(int i = 1; i <= n; i++)
			for(int j =0;j<=v; j++) //注意要从0开始，这个题测试数据有点变态，有的骨头有价值，但占的空间是0
			{
			  if(j-vol[i]>=0)
				  dp[i][j] = max(dp[i-1][j], dp[i-1][j-vol[i]]+val[i]);
			  else
				  dp[i][j]=dp[i-1][j];
			}
			printf("%d\n",dp[n][v]);
	}
	return 0;
}

//一维数组实现01背包


#include<cstdio>
#include<cstring>

#define N 1005
#define max(a, b)  (a) > (b) ? (a) : (b)

int val[N];
int vol[N];
int dp[N];

int main ()
{
	int T, n, v;
	scanf("%d", &T);
	while(T--)
	{
		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]);

		memset(dp,0,sizeof(dp));
		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\n", dp[v]);
	}
	return 0;
}