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hdu 2602 Bone Collector

2024-07-12 05:35:16   228人阅读

简单的01背包，题意很清晰啊。

对于背包问题我有一个建议就是都做题。

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 27979    Accepted Submission(s): 11404


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 
/*
问题分析：第二行输入骨头的value[1001]，第二行输入骨头的volume[1001]
我们要得到状态方程dp[i][j]=max(dp[i-1][j],dp[i-1][j-volume[i]]+value[i])
方程解释：当装入第i个骨头时我们应该考虑到装入第i-1个骨头的价值dp[i][j],
我们要比较装入第i与没有装第i个的价值的比较。没有装入第i个就是将第i-1个装入体积为j的包中
当装入第i个时就是把第i-1个装入体积为j-volume[i]的包裹中，但是由于装入了第i个所以它的的价值增加了value[i];
*/
#include<iostream>
#include<cstring>
using namespace std;
int max(int a,int b)
{
  if(a>b)
	  return a;
  else
	  return b;
}
int main(int i,int j)
{
  int t;//测试样例
  int n;//总骨头数目
  int v;//总的包的体积
  int static dp[1001][1001];//最后的最大价值
  int value[1001];//单个骨头的价值
  int volume[1001];//单个骨头的体积
  cin>>t;
  while(t--)
  {
    cin>>n>>v;
	for(i=1;i<=n;i++)
		cin>>value[i];
	for(i=1;i<=n;i++)
		cin>>volume[i];
	memset(dp,0,sizeof(dp));
	for(i=1;i<=n;i++)
	{
	  for(j=0;j<=v;j++)
	  {
		  if(j>=volume[i])
	    dp[i][j]=max(dp[i-1][j],dp[i-1][j-volume[i]]+value[i]);
		  else
			  dp[i][j]=dp[i-1][j];
	  }
	}
	cout<<dp[n][v]<<endl;
  }
  return 0;
}

 


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