分割槽問題
對於這個問題,給定的集合可以分部分解,每個部分的和相等。
首先,我們要找出給定集合的和。如果它是偶數,那麼有機會將它分成兩組。否則,不能被分割。
對於和的偶數值,我們將建立一個名為 partTable 的表,現在使用以下條件來解決問題。
當陣列 [0] 到陣列 [j-1] 的子集的和等於 i 時,partTable[i, j] 為真,否則為假。
輸入和輸出
Input: A set of integers. {3, 1, 1, 2, 2, 1} Output: True if the set can be partitioned into two parts with equal sum. Here the answer is true. One pair of the partitions are: {3, 1, 1}, {2, 2, 1}
演算法
checkPartition(set, n)
輸入 −給定集合,集合中的元素個數。
輸出 − 當可以分隔時為真,形成兩個和相等的子集。
Begin sum := sum of all elements in the set if sum is odd, then return define partTable of order (sum/2 + 1 x n+1) set all elements in the 0th row to true set all elements in the 0th column to false for i in range 1 to sum/2, do for j in range 1 to n, do partTab[i, j] := partTab[i, j-1] if i >= set[j-1], then partTab[i, j] := partTab[i, j] or with partTab[i – set[j-1], j-1] done done return partTab[sum/2, n] End
示例
#include <iostream> using namespace std; bool checkPartition (int set[], int n) { int sum = 0; for (int i = 0; i < n; i++) //find the sum of all elements of set sum += set[i]; if (sum%2 != 0) //when sum is odd, it is not divisible into two set return false; bool partTab[sum/2+1][n+1]; //create partition table for (int i = 0; i <= n; i++) partTab[0][i] = true; //for set of zero element, all values are true for (int i = 1; i <= sum/2; i++) partTab[i][0] = false; //as first column holds empty set, it is false // Fill the partition table in botton up manner for (int i = 1; i <= sum/2; i++) { for (int j = 1; j <= n; j++) { partTab[i][j] = partTab[i][j-1]; if (i >= set[j-1]) partTab[i][j] = partTab[i][j] || partTab[i - set[j-1]][j-1]; } } return partTab[sum/2][n]; } int main() { int set[] = {3, 1, 1, 2, 2, 1}; int n = 6; if (checkPartition(set, n)) cout << "Given Set can be divided into two subsets of equal sum."; else cout << "Given Set can not be divided into two subsets of equal sum."; }
輸出
Given Set can be divided into two subsets of equal sum.
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