HDOJ 2602 Bone Collector 0-1背包问题的最简形

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 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.

One integer per line representing the maximum of the total value (this number will be less than 2 ³¹).

  
1
5 10
1 2 3 4 5
5 4 3 2 1

  
14
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#include <cctype>
#define N 1010
#define Max 0xffff
int dp[N][N];
//表示取前i种物品，使它们的总体积不超过j的最优取法取得的价值总和
using namespace std;
int main()
{int cas,n,V;cin>>cas;while(cas--){memset(dp,0,sizeof(dp));int d[N] {0},w[N] {0};/*用两个数组分别存储价值和体积*/cin>>n>>V;for(int i=1; i<=n; ++i)cin>>d[i];for(int i=1; i<=n; i++)cin>>w[i];for(int i=1; i<=n; i++){for(int j=0; j<=V; j++){if(w[i]<=j)dp[i][j]=max(dp[i-1][j],dp[i-1][j-w[i]]+d[i]);//对于每一件物品，有放或者不放两种选择(如果能放下)elsedp[i][j]=dp[i-1][j];}}cout<<dp[n][V]<<endl;}return 0;
}