01背包

01背包解决一些有一定代价和一定价值的物品，求解所给你所有代价所能取得物品的最多价值，并且一个物品只能取一次。

使用的思想是dp，每步取最优解，每步都有且只有两种情况，取或者不取这两种情况。查找状态方程，f[i][j]=max(f[i-1][j-c[i]]+w[i],f[i-1][j-c[i]]);

该方程中j表示物品的代价，对于每个j而言，都是取能从物品中选择的最大价值，一次累加，直到最后（自己所持有的代价），而i则表示每次放入一个物品，控制放置物品的个数。因此方程也可以转化，改变空间复杂度，即f[j]=max（f[j-c[i]]+w[i],f[j-c[i]]）;

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 一道简单的01背包问题，可以根据01背包的状态方程，可以求解出答案。 （若能把状态方程的执行自己按步骤执行出来，可以更好理解01背包） 代码：

#include<stdio.h>
#include<string.h>
int max(int a,int b)
{
	return a>b?a:b;
}
int main()
{
	int i,k,t,m,n,v,j;
	int a[1010];
	int c[1010];
	int w[1010];
	scanf("%d",&t);
	while(t--)
	{
	  memset(a,0,sizeof(a));
	  memset(c,0,sizeof(c));
	  memset(w,0,sizeof(w));
      scanf("%d%d",&n,&v);
	  for(i=1;i<=n;i++)
	   scanf("%d",&w[i]);
	  for(i=1;i<=n;i++)
	    scanf("%d",&c[i]);
	  for(i=1;i<=n;i++)
	  {
	  	 for(k=v;k>=c[i];k--)
	  	{
	  	   a[k]=max(a[k-c[i]]+w[i],a[k]);
			  }		
	   }
	printf("%d\n",a[v]);
    }
    return 0;
}

其中内循环中k从v到从c[i]，可以保证每次每个物品只取一次，该处可以和完全背包相比较理解。模拟状态方程执行情况，更容易理解。