hdu 2602 * 01背包 一维数组 不限装满

 

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 2³¹).

 

Sample Input

15 101 2 3 4 55 4 3 2 1

 

Sample Output

14

 

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;}