144 lines
4.6 KiB
C++
Executable File
144 lines
4.6 KiB
C++
Executable File
/***************************************************************************** *
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* Copyright (C) 2022 Mingtian Shao shaomt@nudt.edu.cn *
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* *
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* This file is part of homework of Algorithm Design and Analysis. *
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* *
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* @file algorithm.cpp *
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* @brief Implement different algorithms *
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* *
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* @author Mingtian Shao *
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* @email shaomt@nudt.edu.cn *
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* @version 1.0 *
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* @date 2022-11-12 *
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* @license GNU General Public License (GPL) *
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* *
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*----------------------------------------------------------------------------*
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* Change History : *
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* <Date> | <Version> | <Author> | <Description> *
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*----------------------------------------------------------------------------*
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* 2022/11/12 | 1.0 | Mingtian Shao | Create file *
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*----------------------------------------------------------------------------*
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* *
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*****************************************************************************/
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#include <iostream>
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#include <vector>
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#include <ctime>
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#include <cstdlib>
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#include <stack>
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#include <string>
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#include <cstring>
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#include "algorithm.h"
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using namespace std;
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/**
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* @func algorithm_1
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* @brief 解决背包问题思路一
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* @param n: 物品种类
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* @param w: 背包容量限制
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* @param weights: 物品重量数组
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* @param values: 物品价值数组
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* @param solve: 保存解的数组
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*
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* @return 返回背包容纳的最大价值
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*/
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int algorithm_1(int n, int w, int* weights, int* values, int* solve)
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{
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// memo index start from 0, which means type-1 or weight-1
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vector< vector<int> > memo(n);
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vector< vector<int> > nums(n);
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for (int i = 0; i < n; i++)
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{
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memo[i].resize(w);
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nums[i].resize(w);
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}
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// avoid the inefficient branch for i = 0 in the following ciculation
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for (int j = 0; j < w; j++) // avaliable weight capacity - 1
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{
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int k = (j + 1) / weights[0];
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memo[0][j] = k * values[0]; // take as many as possible
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nums[0][j] = k;
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}
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for (int i = 1; i < n; i++) // index of chosen object
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{
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for (int j = 0; j < w; j++) // avaliable weight capacity - 1
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{
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int maxvalue = 0, maxk = 0;
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for (int k = 0; k <= (j + 1) / weights[i]; k++)
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{
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int tmp = k * values[i];
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if (j - k * weights[i] >= 0)
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{
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tmp += memo[i - 1][j - k * weights[i]];
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}
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if (tmp > maxvalue)
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{
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maxvalue = tmp;
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maxk = k;
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}
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}
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memo[i][j] = maxvalue;
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nums[i][j] = maxk;
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}
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}
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int tmpw = w - 1;
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for (int i = n - 1; i >= 0; i--)
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{
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solve[i] = nums[i][tmpw];
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tmpw -= solve[i] * weights[i];
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}
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return memo[n - 1][w - 1];
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}
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/**
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* @func 函数名
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* @brief 函数简要说明
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* @param 参数1
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* @param 参数2
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*
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* @return 返回说明
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*/
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int algorithm_2(int n, int w, int* weights, int* values, int* solve)
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{
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/*
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* coding
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*/
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}
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/**
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* @func algorithm_1_analy
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* @brief 在函数algorithm_1的基础上进行插桩，记录关键操作次数
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* @param n: 物品种类
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* @param w: 背包容量限制
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* @param weights: 物品重量数组
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* @param values: 物品价值数组
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* @param solve: 保存解的数组
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* @param calc_count: 记录计算次数
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* @param search_count: 记录查找次数
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*
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* @return 返回背白容纳的最大价值
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*/
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int algorithm_1_analy(int n, int w, int* weights, int* values, int* solve, int& calc_count, int& search_count)
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{
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/*
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* coding
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*/
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}
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/**
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* @func 函数名
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* @brief 函数简要说明
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* @param 参数1
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* @param 参数2
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*
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* @return 返回说明
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*/
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int algorithm_2_analy(int n, int w, int* weights, int* values, int* solve, int& calc_count, int& search_count)
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{
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/*
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* coding
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*/
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} |