HDU2602

来源:互联网 发布:mac删除双系统 编辑:程序博客网 时间:2024/05/07 17:22
题目的链接:http://acm.hdu.edu.cn/showproblem.php?pid=2602
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
简单的01背包,但是要注意数组的大小和输入的数据,注意 f[0]有价值。
#include<cstdio>#include<iostream>#include<cstring>#include<algorithm>using namespace std;int V[1010];int w[1010];int f[1010];int main(){    int T;    scanf("%d",&T);    while( T--)    {        ///memset(V,0,sizeof(V));        ///memset(w,0,sizeof(w));        int n,v;        scanf("%d%d",&n,&v);        for(int i=1;i<=n;i++)        {            scanf("%d",&w[i]);        }        for(int i=1;i<=n;i++)        {            scanf("%d",&V[i]);        }        memset(f,0,sizeof(f));        for(int i=1;i<=n;i++)        {            for(int j=v;j>=V[i];j--)            {                f[j]=max(f[j],f[j-V[i]]+w[i]);            }        }        printf("%d\n",f[v]);    }    return 0;}



0 0