初级背包模板

0-1背包

Bone Collector

Time Limit: 1000MS Memory Limit: 65536KB

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).

 

Example Input

15 101 2 3 4 55 4 3 2 1

Example Output

14

Hint

hdoj2602

Author

#include <iostream>#include<bits/stdc++.h>using namespace std;int main(){    int t;    cin>>t;    while(t--)    {    int n,v;    scanf("%d%d",&n,&v);    int g[3000]={0},w[3000],tj[3000];    for(int i=1;i<=n;i++)    scanf("%d",&w[i]);    for(int j=1;j<=n;j++)    scanf("%d",&tj[j]);        for(int i=1; i<=n; i++)        {            for(int j=v; j>=1; j--) //j代表当前的背包容量            {                if(tj[i]<=j)//假如第i件物品可以放进                {                    g[j] = max(g[j],g[j-tj[i]]+w[i]);//腾出背包，并依据其价值判断是否放进                }            }        }    printf("%d\n",g[v]);    }    return 0;}