在这个问题中,给出了一组整数,我们必须将它们分为两部分,以使两个子集的和之差尽可能小。因此,我们的目标是将力量几乎相等的两组分成参加拔河比赛。
如果子集n的大小为偶数,则必须将其分为n / 2,但对于n的奇数,则一个子集的大小必须为(n-1)/ 2,而另一个子集的大小必须为( n + 1)/ 2。
Input: A set of different weights. {23, 45, -34, 12, 0, 98, -99, 4, 189, -1, 4} Output: The left and right subset to distribute the weights to make the difference minimum. Left: {45, -34, 12, 98, -1} Right: {23, 0, -99, 4, 189, 4}
tugOfWar(weight, n, curr, select, sol, diff, sum, total, pos)
输入- 给定权重,权重数量,当前列表,所选项目的数量,两个子集总和之间的差,所有项目的总和,子集中的总和,所选元素的位置的集合。
输出- 为左右子集选择的解决方案集。
Begin if pos = n, then //when all elements are taken return if (n/2-select) > (n - pos), then return tugOfWar(weight, n, curr, select, sol, diff, sum, total, pos+1) select := select + 1 total := total + weight[pos] curr[pos] := true //when item at pos is taken if select = n/2, then if difference of (sum/2 and total) < diff, then diff := difference of (sum/2 and total) for i := 0 to n, do sol[i] := curr[i] done else tugOfWar(weight, n, curr, select, sol, diff, sum, total, pos+1) curr[pos] := false //remove current element if not properly done End
#include <iostream> #include<cmath> using namespace std; void tugOfWar(int* weight, int n, bool curr[], int select, bool sol[], int &diff, int sum, int total, int pos) { if (pos == n) //when the pos is covered all weights return; if ((n/2 - select) > (n - pos)) //left elements must be bigger than required result return; tugOfWar(weight, n, curr, select, sol, diff, sum, total, pos+1); select++; total += weight[pos]; curr[pos] = true; //add current element to the solution if (select == n/2) { //when solution is formed if (abs(sum/2 - total) < diff) { //check whether it is better solution or not diff = abs(sum/2 - total); for (int i = 0; i<n; i++) sol[i] = curr[i]; } } else { tugOfWar(weight, n, curr, select, sol, diff, sum, total, pos+1); } curr[pos] = false; //when not properly done, remove current element } void findSolution(int *arr, int n) { bool* curr = new bool[n]; bool* soln = new bool[n]; int diff = INT_MAX; //set minimum difference to infinity initially int sum = 0; for (int i=0; i<n; i++) { sum += arr[i]; //find the sum of all elements curr[i] = soln[i] = false; //make all elements as false } tugOfWar(arr, n, curr, 0, soln, diff, sum, 0, 0); cout << "Left: "; for (int i=0; i<n; i++) if (soln[i] == true) cout << arr[i] << " "; cout << endl << "Right: "; for (int i=0; i<n; i++) if (soln[i] == false) cout << arr[i] << " "; } int main() { int weight[] = {23, 45, -34, 12, 0, 98, -99, 4, 189, -1, 4}; int n = 11; findSolution(weight, n); }
输出结果
Left: 45 -34 12 98 -1 Right: 23 0 -99 4 189 4