题目：
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 ?
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.
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
有一个收集骨头的人，现在有n种骨头，背包总容量为v，现在给出n和v，以及ni种骨头的价值以及体积，求所能收集到的最大值。
思路：
01背包问题。
源代码：
#include <bits/stdc++.h>
using namespace std;
int main()
{
int N;
int n,m,i,t;
cin >> N;
int V[10005],value[10005];
int dp[10005];
while (N--)
{
cin >> n >> m;
for (i = 1; i <= n; i++)
cin >> value[i];
for (i = 1; i <= n; i++)
cin >> V[i];
memset(dp,0,sizeof(dp));
for (i = 1; i <= n; i++)
for (t = m; t >= V[i] ; t--)
{
if (dp[t] < dp[t - V[i]]+value[i])
dp[t] = dp[t - V[i]]+value[i];
}
cout << dp[m] << endl;
}
return 0;
}