2018-BNUZ-ACM-1(背包) HDU-2602

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 76935    Accepted Submission(s): 31880

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

15 101 2 3 4 55 4 3 2 1

 

Sample Output

14

 

Author

Teddy

 

Source

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

 

Recommend

lcy

#include
#include
#include
using namespace std;

int dp[1005][1005];

int main()
{
	int v,m,t;
	int weight[1005];
	int value[1005];
	cin >> t;
	while(t--)
	{
		cin >> m >> v;
		memset(dp,0,sizeof(dp));
		for(int i = 1 ; i <= m ; i++)
		{
			cin >> value[i];
		}
		for(int i = 1 ; i <= m ; i++)
		{
			cin >> weight[i];
		}
		for(int i = 1 ; i <= m ; i++)
		{
			for(int j = 0 ; j <= v ; j++)
			{
				if(j >= weight[i])
				{
					dp[i][j] = max(dp[i-1][j],dp[i-1][j-weight[i]]+value[i]);
				}
				else
				dp[i][j] = dp[i-1][j];
			}
		}
		cout << dp[m][v] << endl;
	}
}

题意是简单的0/1背包。在背包空间一定的情况下尽量装的总价值最高。核心公式

dp[i][j] = max(dp[i-1][j],dp[i-1][j-weight[i]]+value[i]);