Bone Collector（01背包）

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

#include<stdio.h>
#include<string.h>
#define Max 1010
int c[Max],w[Max],V,N,dp[Max];
int main()
{
	//freopen("b.txt","r",stdin);
	int t,i;
	scanf("%d",&t);
	while(t--)
	{
		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]);
		memset(dp,0,sizeof(dp));
		for(i=1;i<=N;i++)
			for(int j=V;j>=c[i];j--)
				dp[j]=dp[j]>dp[j-c[i]]+w[i]?dp[j]:dp[j-c[i]]+w[i];
			printf("%d\n",dp[V]);
	}
	return 0;
}
		

输出选择方案的代码：

#include<stdio.h>
#include<string.h>
#define Max 1010
int c[Max],w[Max],V,N,dp[Max];
int path[Max];
int main()
{
	freopen("b.txt","r",stdin);
	int t,i,pre;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d %d",&N,&V);
		for(i=1;i<=N;i++)
			scanf("%d",&c[i]);
		for(i=1;i<=N;i++)
			scanf("%d",&w[i]);
		memset(dp,0,sizeof(dp));
		memset(path,0,sizeof(path));
		pre=-1;
		for(i=1;i<=N;i++)
			for(int j=V;j>=c[i];j--)
				if(dp[j-c[i]]+w[i]>dp[j])
				{
					dp[j]=dp[j-c[i]]+w[i];
					path[i]=1;
				}
				printf("%d\n",dp[V]);
				for(i=1;i<=N;i++)
					if(path[i]) printf("%d ",i);
					printf("\n");
	}
	return 0;
}