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Bone CollectorTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) 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
翻译：有一人爱收集骨头，每个骨头有一个价值和重量，他有一个背包，容量为V，问他装得最大价值量是多少？（经典的01背包问题）
重点：01背包、动态规划
难点：动态方程的书写
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int main()
{
int T,n,v,i,j,dp[1100],a[1100],b[1100];//a[1100]是价值，b[1100]是重量；
cin>>T;
while(T--)
{
memset(a,0,sizeof(a));
memset(b,0,sizeof(b));
memset(dp,0,sizeof(dp));
cin>>n>>v;//n是骨头个数，V是背包容量
for(i=1;i<=n;i++)
scanf("%d",&a[i]);
for(i=1;i<=n;i++)
scanf("%d",&b[i]);
for(i=1;i<=n;i++)
for(j=v;j>=b[i];j--)
{
if(dp[j]<dp[j-b[i]]+a[i])
dp[j]=dp[j-b[i]]+a[i];
}
printf("%d\n",dp[v]);
}
return 0;
}
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