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01背包:HDU2602Bone Collector
阅读量:3899 次
发布时间:2019-05-23

本文共 1565 字,大约阅读时间需要 5 分钟。

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

题意:

第一行输入例子的组数,然后接着三行
第二行分别输入N,V代表骨头的数量和包的容量
第三行输入每个骨头的价值
第四行输入每个骨头的容量

#include
#include
#include
using namespace std;//HDU2602Bone Collectorint main(){
int z,N,V; int i,j; int f[100]={
0}; int vg[100]={
0}; int rg[100]={
0}; while(scanf("%d",&z)!=EOF){
//输入组数 scanf("%d%d",&N,&V); //再依次输入骨头价值和容量 for(i=0;i
=rg[i];j--)//逆序表达 f[j]=max(f[j],f[j-rg[i]]+vg[i]); printf("%d\n",f[V]); }}

转载地址:http://mkfen.baihongyu.com/

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