我有n
袋糖果,这样两个袋子里面没有相同数量的糖果(即它是A[] = {a0,a1,a2,...,ai,...,aj}
ai != aj
所在的集合。{/ p>
我知道每个袋子里有多少个糖果和我拥有的糖果总数M
。
我需要将这些行李分成三个孩子,以便尽可能公平地分发糖果(即每个孩子尽可能接近M/3
)。
毋庸置疑,我可能不会撕掉袋子来平衡计数 - 那么问题就是微不足道。
有没有人有任何想法如何解决这个问题 - 最好是用Java?
编辑:
面试官希望我使用二维数组来解决问题:第一个孩子得到x,第二个孩子得到y,第三个得到其余的孩子:S[x][y]
。
这是我尝试过之后:
1] sort array n lg n
2] starting with largest remaining bag, give bag to kid with fewest candy.
这是我分区给两个孩子的解决方案(这是正确的答案)。也许它有助于获得3路分区。
int evenlyForTwo(int[] A, int M) {
boolean[] S = new boolean[M+1];
S[0]=true;//empty set
for(int i=0; i<A.length; i++)
for(int x=M; x >= A[i]; x--)
if(!S[x])
S[x]=S[x-A[i]];
int k = (int) M/2;
while(!S[k])
k--;
return k;//one kid gets k the other the rest.
}//
答案 0 :(得分:3)
您描述的问题称为3-Partition problem,并且已知为NP难。在MathOverflow上讨论了这个问题。你可能会发现一些有一些价值的指针。
答案 1 :(得分:1)
这是一个小解决方案,粗略但是给出了正确的结果。你甚至可以改变孩子,包等的数量。
public class BagOfCandies {
static public void main(String...args) {
int repeat = 10;
int childCount = 3;
int bagsCount = childCount + (int) (Math.random() * 10);
for (int k=0; k<repeat; k++) {
int candyCount = 0, n=0;
int[] bags = new int[bagsCount];
for (int i=0; i<bags.length; i++) {
n += 1 + (int) (Math.random() * 2);
bags[i] = n;
candyCount += n;
}
shuffle(bags); // completely optional! It works regardless
boolean[][] dist = divideBags(bags, childCount);
System.out.println("Bags of candy : " + Arrays.toString(bags) + " = " + bags.length);
System.out.println("Total calculated candies is " + candyCount);
int childCandySum = 0;
for (int c=0; c<childCount; c++) {
int childCandies = countSumBags(bags, dist[c]);
System.out.println("Child " + (c+1) + " = " + childCandies + " --> " + Arrays.toString(dist[c]));
childCandySum += childCandies;
}
System.out.println("For a total of " + childCandySum + " candies");
System.out.println("----------------");
}
}
static private void shuffle(int[] bags) {
for (int i=0, len=bags.length; i<len; i++) {
int a = (int)Math.floor(Math.random()*len);
int b = (int)Math.floor(Math.random()*len);
int v = bags[a];
bags[a] = bags[b];
bags[b] = v;
}
}
static private boolean[][] divideBags(int[] bags, int childCount) {
int bagCount = bags.length;
boolean[][] dist = new boolean[childCount][bagCount];
for (int c=0; c<childCount; c++)
Arrays.fill(dist[c], false);
for (int i=0; i<bagCount; i+=childCount)
for (int j=i, c=0; c<childCount && j<bagCount; j++, c++)
dist[c][j] = true;
if (childCount == 1) return dist; // shortcut here
int sumDiff = 1;
int oldDiff = 0;
while (sumDiff != oldDiff) {
oldDiff = sumDiff;
sumDiff = 0;
// start comparing children in pair
for (int child1=0; child1<childCount-1; child1++) {
for (int child2=child1+1; child2<childCount; child2++) {
int count1 = countSumBags(bags, dist[child1]);
int count2 = countSumBags(bags, dist[child2]);
int diff = Math.abs(count1 - count2);
// a difference less than 2 is negligeable
if (diff > 1) {
// find some bags with can swap to even their difference
int c1=-1, c2=-1, cdiff;
boolean swap = false;
for (int i=0; i<bagCount-1; i++) {
for (int j=i; j<bagCount; j++) {
if (dist[child1][i] && dist[child2][j]) {
cdiff = Math.abs((count1 - bags[i] + bags[j]) - (count2 + bags[i] - bags[j]));
if (cdiff < diff) {
c1 = i; c2 = j;
diff = cdiff;
swap = true;
}
}
if (dist[child1][j] && dist[child2][i]) {
cdiff = Math.abs((count1 - bags[j] + bags[i]) - (count2 + bags[j] - bags[i]));
if (cdiff < diff) {
c1 = j; c2 = i;
diff = cdiff;
swap = true;
}
}
}
}
if (swap) {
//System.out.println("Swaping " + c1 + " with " + c2);
dist[child1][c1] = false; dist[child1][c2] = true;
dist[child2][c1] = true; dist[child2][c2] = false;
}
}
//System.out.println("Diff between " + child1 + "(" + countSumBags(bags, dist[child1]) + ") and " + child2 + "(" + countSumBags(bags, dist[child2]) + ") is " + diff);
sumDiff += diff;
}
}
//System.out.println("oldDiff="+oldDiff+", sumDiff="+sumDiff);
}
return dist;
}
static private int countSumBags(int[] bags, boolean[] t) {
int count = 0;
for (int i=0; i<t.length; i++) {
if (t[i]) {
count+=bags[i];
}
}
return count;
}
}
我不知道你是否正在寻找这个结果,但似乎是我对这个问题的理解。