关键字：01背包

题目：
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

链接：
http://acm.hdu.edu.cn/showproblem.php?pid=2602

代码：

#include <iostream>
#include <fstream>
#include <cmath>
using namespace std;
const int maxn = 1010;
const int maxv = 1010;

int main(){
//    freopen("a.txt", "r", stdin);
    int t, n, V;
    int dp[maxn][maxv];
    int w[maxn], v[maxn];

    cin >> t;
    while(t--){
        cin >> n >> V;
        for(int i = 1; i <= n; ++i){
            cin >> w[i];
        }
        for(int i = 1; i <= n; ++i){
            cin >> v[i];
        }
        memset(dp, 0, sizeof(dp));
        for(int i = 1; i <= n; ++i){
            for(int j = 0; j <= V; j++){

                if(j >= v[i]){
                    dp[i][j] = dp[i-1][j-v[i]]+w[i] > dp[i-1][j] ? dp[i-1][j-v[i]]+w[i] : dp[i-1][j];
                }
                else{
                    dp[i][j] = dp[i - 1][j];
                }

            }
        }
        cout << dp[n][V] << endl;

    }
    return 0;
}