题目:
n 支队伍比赛,分别编号为0,1,2......n-1,已知它们之间的实力对比关系,存储在一个二维数组w[n][n]中,w[i][j] 的值代表编号为i,j 的队伍中更强的一支,所以w[i][j]=i 或者j,现在给出它们的出场顺序,并存储在数组order[n]中,比如order[n] = {4,3,5,8,1......},那么第一轮比赛就是4 对3, 5 对8。然胜者晋级,败者淘汰,同一轮淘汰的所有队伍排名不再细分,即可以随便排,下一轮由上一轮的胜者按照顺序,再依次两两比,比如可能是4 对5,直至出现第一名编程实现,给出二维数组w,一维数组order
和用于输出比赛名次的数组result[n],求出result。
代码如下:
/** * @author PLA n支队伍比赛,微软T36 */ public static void main(String[] args) { int[] order = { 3, 2, 5, 1, 4, 6, 8, 7 }; int[][] w = { { 0, 0, 0, 0, 0, 0, 0, 0, 0 }, { 0, 0, 1, 1, 4, 5, 6, 1, 8 }, { 0, 1, 0, 2, 2, 5, 6, 7, 2 }, { 0, 1, 2, 0, 3, 3, 6, 7, 3 }, { 0, 4, 2, 3, 0, 4, 6, 7, 4 }, { 0, 5, 5, 3, 4, 0, 5, 7, 8 }, { 0, 6, 6, 6, 6, 5, 0, 6, 8 }, { 0, 1, 7, 7, 7, 7, 6, 0, 8 }, { 0, 8, 2, 3, 4, 8, 8, 8, 0 } }; LinkedList<Integer> order_List = new LinkedList<Integer>(); for(int m=0;m<order.length;m++){ order_List.add(order[m]); } System.out.println("队伍出场顺序:" + "\n" + order_List); compare(order, w); } private static LinkedList<Integer> rank = new LinkedList<Integer>(); private static Object compare(int[] order, int[][] w) { // TODO Auto-generated method stub int length = order.length; LinkedList<Integer> list = new LinkedList<Integer>(); for (int i = 0; i < length; i += 2) { list.add(w[order[i]][order[i + 1]]); rank.add(w[order[i]][order[i + 1]] == order[i] ? order[i + 1] : order[i]); } ; int size = list.size(); int[] arr = new int[size]; for (int j = 0; j < list.size(); j++) { arr[j] = list.get(j); } if (size == 1) { rank.add(list.get(0)); System.out.println("排名:" + "\n" + rank_Reverse(rank)); return null; } return compare(arr, w); } private static LinkedList<Integer> rank_Reverse(LinkedList<Integer> rank2) { // TODO Auto-generated method stub int temp = 0; for (int i = 0, j = rank2.size() - 1; i < j; i++, j--) { temp = rank2.get(i); rank2.set(i, rank2.get(j)); rank2.set(j, temp); } return rank2; }