HDU 2602 Bone Collector

最新推荐文章于 2021-08-01 08:53:01 发布

良月澪二

最新推荐文章于 2021-08-01 08:53:01 发布

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分类专栏： HDU 背包问题背包问题深入讲解文章标签：动态规划背包 NOIP

本文链接：https://blog.youkuaiyun.com/yandaoqiusheng/article/details/84846409

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本文讲解了一个经典的计算机科学问题——01背包问题。通过一个具体的案例，即骨收集者的最大价值问题，详细介绍了如何使用动态规划算法来解决该问题。文章提供了完整的C++代码实现，帮助读者理解算法的具体应用。

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题目链接：传送门
题面：
$B o n e C o l l e c t o r$

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 ?
在这里插入图片描述

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

题目大意

输入多组数据T。你有一个容量为 $V$ 的背包，有许多价值和体积不等的骨头，你要用背包装下他们其中的一些骨头使价值最大。第一行输入骨头数量 $n$ 和背包体积 $V$ ，第二行有 $n$ 个数，是每个骨头的价值，第三行有 $n$ 个数，是每个骨头的体积。输出有 $T$ 行，是你求得的最大价值。

这
板子啊
01背包

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <complex>
#include <algorithm>
#include <climits>
#include <queue>
#include <map>
#include <vector>
#include <iomanip>
#define A 1000010
#define B 2010
#define ll long long

using namespace std;
int T, n, V, w[A], v[A], f[A];

int main() {
    cin >> T;
    while (T--) {
        memset(f, 0, sizeof f);
        cin >> n >> V;
        for (int i = 1; i <= n; i++) cin >> v[i];
        for (int i = 1; i <= n; i++) cin >> w[i];
        for (int i = 1; i <= n; i++)
          for (int j = V; j >= w[i]; j--)
            f[j] = max(f[j], f[j - w[i]] + v[i]);
        cout << f[V] << endl;
    }
}