hdu 2602 Bone Collector（我的第一个01背包）

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7904    Accepted Submission(s): 2949

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

 

     这题就是最简单的01背包，逆序！！！

      for i=1..N

              for v=V..0

                  f[v]=max{f[v],f[v-c[i]]+w[i]};

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

代码：

#include <iostream>
#include <stdio.h>
#include <memory.h>
using namespace std;

int c[1005], w[1005];
int bag[1005];
int N, V;

void _01_bag()  //01背包：逆序！
{
    int i, j;
    memset(bag, 0, sizeof(bag));
    for(i = 0; i < N; i++)
    {
        for(j = V; j >= c[i]; j--)
        {
            bag[j] = max(bag[j], bag[j-c[i]] + w[i]);
        }
    }
}

int main()
{
    int i, t;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d %d", &N, &V);
        for(i = 0; i < N; i++)
            scanf("%d", &w[i]);
        for(i = 0; i < N; i++)
            scanf("%d", &c[i]);
        _01_bag();
        printf("%d\n", bag[V]);
    }

    return 0;
}