HDU2602:Bone Collector

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

用子问题定义状态：即dp[i][v]表示前i件物品恰放入一个容量为v的背包可以获得的最大价值。则其状态转移方程便是： dp[i][v]=max{dp[i-1][v],dp[i-1][v-vi[i]]+val[i]} 这个方程非常重要，基本上所有跟背包相关的问题的方程都是由它衍生出来的。所以有必要将它详细解释一下： “将前i件物品放入容量为v的背包中”这个子问题 若只考虑第i件物品的策略（放或不放），那么就可以转化为一个只牵扯前i-1件物品的问题。 如果不放第i件物品，那么问题就转化为“前i-1件物品放入容量为v的背包中”，价值为dp[i-1][v]； 如果放第i件物品，那么问题就转化为“前i-1件物品放入剩下的容量为v-vi[i]的背包中”，此时能获得的最大价值就是dp[i-1][v-vi[i]]再加上通过放入第i件物品获得的价值val[i]。

#include<cstdio>
#include<cstring>
#include<algorithm> 
using namespace std;
int dp[1000][1001];
int  vi[10001];
int   val[10001];
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",&vi[i]);
		}
		dp[0][0]=0;
		for(int i=1;i<=n;i++)
		{
				for(int j=0;j<=v;j++)
	    	{
	    		if(vi[i]<=j)
			    dp[i][j]=max(dp[i-1][j],dp[i-1][j-vi[i]]+val[i]);//放入第i个物品 
			    else
			    dp[i][j]=dp[i-1][j];
		    }
		}
		printf("%d
",dp[n][v]);
	}
	return 0;
}

下面是对上面的优化

我们可以用一维数组来代替二维数组

dp[i][v]是由dp[i-1][v]和dp[i-1][v-vi[i]]两个子问题递推而来，为了保证在推dp[i][v]时（也即在第i次主循环中推dp[v]时）能够得到dp[i-1][v]和dp[i-1][v-vi[i]]的值 可以在每次i主循环中以v——>0求出dp[v]

状态转移方程：dp[v]=max(dp[v],dp[v-vi[i]+val[i]);

注意：这种解法只能由V--0，不能反过来，如果反过来就会造成物品重复放置！

#include<cstdio>
#include<cstring>
#include<algorithm> 
using namespace std;
int dp[1000];
int  vi[10001];
int   val[10001];
int main()
{
	int t; 
	scanf("%d",&t);
	while(t--)
	{	
	   memset(dp,0,sizeof(dp));
		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",&vi[i]);
		}
		dp[0]=0;
		for(int i=1;i<=n;i++)
		{
				for(int j=v;j>=vi[i];j--)
	    	{
	    	
			    dp[j]=max(dp[j],dp[j-vi[i]]+val[i]);
			  
		    }
		}
		printf("%d
",dp[v]);
	}
	return 0;
}