77 lines
2.5 KiB
C++
77 lines
2.5 KiB
C++
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// Rod Cutting Problem
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// Given a rod of length n and a list of rod prices of length i, where 1 <= i <= n, find the optimal way to cut the rod into smaller rods to maximize profit.
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// Rod Cutting Optimal Approach
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// We will solve this problem in a bottom-up manner. (iteratively)
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// In the bottom-up approach, we solve smaller subproblems first, then move on to larger subproblems.
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// The following bottom-up approach computes dp[i], which stores maximum profit achieved from the rod of length i from 1 to len.
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// It uses the value of smaller values i already computed.
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// Space complexity: O(n)
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// Time complexity: O(n^n)
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// Solution
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#include <iostream>
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#include <vector>
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#include <climits>
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using namespace std;
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// Function to find the maximum revenue from cutting a rod of length (len)
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// where the rod of length (i) has cost (prices[i - 1])
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int RodCutting(vector<int> &prices, int len)
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{
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// (dp) stores the maximum revenue achieved from cutting a rod of length (from 1 to len)
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vector<int> dp(len + 1, 0);
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// If the rod length is negative (invalid) or zero there's no revenue from it
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if (len <= 0)
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{
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return 0;
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}
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// Cut a rod of length (i)
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for (int i = 1; i <= len; i++)
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{
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// divide the rod of length (i) into two rods of lengths (j) and (i - j)
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// and store the maximum revenue
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for (int j = 0; j < i; j++)
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{
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// (dp[i]) stores the maximum revenue achieved from cutting a rod of length (i)
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dp[i] = max(dp[i], prices[j] + dp[i - j - 1]);
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}
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}
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// (dp[len]) contains the maximum revenue from cutting a rod of length (len)
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return dp[len];
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}
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int main()
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{
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int len;
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cout << "Enter the rod length :";
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cin >> len;
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vector<int> prices(len);
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for (int i = 1; i <= len; i++)
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{
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cout << "Enter the price of the rod of length " << i << " :";
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cin >> prices[i - 1];
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}
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cout << "Maximum revenue = " << RodCutting(prices, len);
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return 0;
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}
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// Input and output:
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// 1. prices[] = [1, 5, 8, 9, 10, 17, 17, 20]
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// rod length = 4
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// Best: Cut the rod into two pieces of length 2 each to gain revenue of 5 + 5 = 10
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// 2. prices[] = [1, 5, 8, 9, 10, 17, 17, 20]
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// rod length = 8
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// Best: Cut the rod into two pieces of length 2 and 6 to gain revenue of 5 + 17 = 22
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// 3. prices[] = [3, 5, 8, 9, 10, 17, 17, 20]
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// rod length = 8
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// Best: Cut the rod into eight pieces of length 1 to gain revenue of 8 * 3 = 24
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