hdu 2602 Bone Collector （简单的01背包）

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

Bone Collector

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

15 101 2 3 4 55 4 3 2 1

Sample Output

14

简单的01背包

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>#include <cmath>#include <cstdlib>#include <limits>#include <queue>#include <stack>#include <vector>#include <map>using namespace std;#define N 1060#define INF 0x3f3f3f3f#define PI acos (-1.0)#define EPS 1e-8#define met(a, b) memset (a, b, sizeof (a))typedef long long LL;int dp[N];int main (){    int t, n, V;    scanf ("%d", &t);    while (t--)    {        int vl[N], vo[N];        scanf ("%d %d", &n, &V);        for (int i=1; i<=n; i++)            scanf ("%d", &vl[i]);        for (int i=1; i<=n; i++)            scanf ("%d", &vo[i]);        met (dp, 0);        for (int i=1; i<=n; i++)            for (int j=V; j>=vo[i]; j--)                dp[j] = max (dp[j], dp[j-vo[i]]+vl[i]);        printf ("%d\n", dp[V]);    }    return 0;}