HDU 2602 Bone Collector

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#hdu #01背包

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本文介绍了一个经典的01背包问题，并给出了使用滚动数组优化空间复杂度的解决方案。问题描述了一个骨骸收集者如何在有限的背包容量下，选择不同体积和价值的骨骸以获得最大总价值。

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

Bone Collector

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 48219 Accepted Submission(s): 20113

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

Sample Output

Author

Teddy

Source

HDU 1st “Vegetable-Birds Cup” Programming Open Contest

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比较经典的01背包。不多解释，这里采用滚动数组优化空间复杂度。

附上AC代码：

#include <bits/stdc++.h>
//#pragma comment(linker, "/STACK:102400000, 102400000")
using namespace std;
const int maxn = 1005;
int dp[maxn];
int c[maxn], w[maxn];
int n, v;

int main(){
	#ifdef LOCAL
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
	#endif
	int T;
	scanf("%d", &T);
	while (T--){
		scanf("%d%d", &n, &v);
		for (int i=0; i<n; ++i)
			scanf("%d", w+i);
		for (int i=0; i<n; ++i)
			scanf("%d", c+i);
		memset(dp, 0, sizeof(int)*(v+1));
		for (int i=0; i<n; ++i)
			for (int j=v; j>=c[i]; --j)
				dp[j] = max(dp[j], dp[j-c[i]]+w[i]);
		printf("%d\n", dp[v]);
	}
	return 0;
}