recursion problem(IsMeasurable)(Is this correct?)

本文介绍了一种递归算法来判断是否能通过给定的一组物品重量组合达到目标重量。通过对物品进行递归分解,算法考虑了每个物品可能被加入、排除或作为负重的情况。该算法适用于解决物品组合称重问题。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

#include <iostream>
#include <iomanip>
#include <string>

using namespace std;

bool IsMeasurable(int target, int weights[], int nWeights);
void SwapArray(int weights[], int i, int j);

int main()
{
	int sampleWeights[] = { 1, 3 };
	int nSampleWeights = 2;

	cout << boolalpha;
	cout << IsMeasurable(2, sampleWeights, nSampleWeights) << endl;
	cout << IsMeasurable(5, sampleWeights, nSampleWeights) << endl;

	return 0;
}

bool IsMeasurable(int target, int weights[], int nWeights)
{
	// simple case
	if (target == 0 && nWeights == 0) {
		return true;
	}
	if ((target == 0 && nWeights != 0) ||
		(target != 0 && nWeights == 0)) {
		return false;
	}

	// recursive decomposition
	for (int i = 0; i < nWeights; i++) {
		int weight = weights[i];
		SwapArray(weights, i, nWeights-1);
		if (IsMeasurable(target-weight, weights, nWeights-1) ||
			IsMeasurable(target, weights, nWeights-1) ||
			IsMeasurable(target + weight, weights,nWeights-1)) {
				return true;
		}
		SwapArray(weights, i, nWeights-1);
	}
	return false;
}

void SwapArray(int weights[], int i, int j) {
	int temp = weights[i];
	weights[i] = weights[j];
	weights[j] = temp;
}

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值