HDU 2602 Bone Collector

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

1

5 10

1 2 3 4 5

5 4 3 2 1

Sample Output

14

题意

01背包问题。
状态转移方程:dp[i][v]=max(dp[i-1][v],dp[i-1][v-cost[i]]+value[i])

代码实现

#include<iostream>
using namespace std;
int dp[1000][1000];

int max(int x,int y)
{
return x>y?x:y;
}

int main()
{
int t,n,v,i,j;
int va[1000],vo[1000];
cin>>t;
while(t--)
{
cin>>n>>v;
for(i=1;i<=n;i++)
cin>>va[i];
for(i=1;i<=n;i++)
cin>>vo[i];
memset(dp,0,sizeof(dp));//初始化操作
for(i=1;i<=n;i++)
{
for(j=0;j<=v;j++)
{
if(vo[i]<=j)//表示第i个物品将放入大小为j的背包中
dp[i][j]=max(dp[i-1][j],dp[i-1][j-vo[i]]+va[i]);//第i个物品放入后，那么前i-1个物品可能会放入也可能因为剩余空间不够无法放入
else //第i个物品无法放入
dp[i][j]=dp[i-1][j];
}
}
cout<<dp
[v]<<endl;
}
return 0;
}