hdu 2602 * 01背包一维数组不限装满

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Bone Collector

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9840 Accepted Submission(s): 3778

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

Sample Output

二维数组：

#include <iostream>
using namespace std;
int opt[1001][1001];
int v[1001],w[1001];

int main()
{
    int t,N,i,j,V;
    cin>>t;
    while(t--)
    {
        cin>>N>>V;
        for(i=1;i<=N;i++)
            scanf("%d",&w[i]);
        for(i=1;i<=N;i++)
            scanf("%d",&v[i]);
        memset(opt,0,sizeof(opt)); //初始化

        for(i=1;i<=N;i++)
            for(j=0;j<=V;j++) //注意要从0开始，这个题测试数据有点变态，有的骨头有价值，但占的空间是0
            {
                if(v[i]<=j && opt[i-1][j]<opt[i-1][j-v[i]]+w[i])
                    opt[i][j]=opt[i-1][j-v[i]]+w[i];
                else
                    opt[i][j]=opt[i-1][j];

            }
        
        cout<<opt[N][V]<<endl;
    }
    return 0;
}

一维数组：

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<iostream>
#define maxV 1111//总价值
#define maxN 1111//总量

long  opt[maxV];
long  w[maxN],v[maxN];
int N,V,t;
int main(void){
	int i,j;
  scanf("%d",&t);
  while(t-->0){
	scanf("%d%d",&N,&V);//V 背包容量
	for(i=0;i<N;i++){
		scanf("%ld",v+i);  //为什么写成v[i]就wa!  但是换成cin是可以的
	}
	for(i=0;i<N;i++){
		scanf("%ld",w+i);
	}
	memset(opt,0,sizeof(opt));//用0初始化 不限定背包装满 即opt[v]不限;
	for(i=0;i<N;i++)
		for(j=V;j-w[i]>=0;j--)   //j-w[i]>=0 即是一维数组里的if判断  也是ZeroOnePark的优化方案 
			opt[j] = std::max(opt[j],opt[j-w[i]]+v[i]);
		
	 printf("%ld\n",opt[V]); 
  }
return 0;
}