多人单循环比赛的安排问题
有5个队伍参加比赛，A，B，C，D，E，每个队都要与其它队互相赛一场，A-B,A-C,A-D,A-E,B-C......而且一个队伍一天只赛一场，如第一天 A-B,C-D,E轮空，怎么用程序把全部赛程排出来？原文：http://www.java2000.net/p11024
<div class="highlighter">
[*]import java.util.HashMap;
[*]<span />import java.util.HashSet;
[*]<span />import java.util.Map;
[*]<span />import java.util.Set;
[*]<span />
[*]<span />/**
[*] * 多人比赛的安排问题。<br>
[*] * 一个人每一天只能赛一场。如果有剩下的则轮空。
[*] *
[*] * @author leoklkj,赵学庆 java2000.net
[*] */<span />
[*]public class T {
[*] static int[] ids = { 1, 2, 3, 4, 5 };
[*]<span />
[*] static Set<Integer> set = new HashSet<Integer>();
[*]<span />
[*] static Set<Integer> setCheck = new HashSet<Integer>();
[*]<span />
[*] static Set<Integer> tempa;
[*]<span />
[*] static Set<Integer> tempb;
[*]<span />
[*] static Map<Integer, Set<Integer>> map = new HashMap<Integer, Set<Integer>>();
[*]<span />
[*] public static void main(String[] args) {
[*] for (int id : ids) {
[*] map.put(id, new HashSet<Integer>());
[*] }
[*] int pointer = 0;
[*] int a;
[*] int b;
[*] int total = ids.length * (ids.length - 1) / 2;
[*] int stop = 20;
[*] while (total > 0 && stop > 0) {
[*] if (pointer >= ids.length) {
[*] pointer = 0;
[*] }
[*] if (pointer < 0) {
[*] pointer = ids.length - 1;
[*] }
[*] setCheck.clear();
[*] a = findNext(pointer);
[*] // 没找到可用的<span />
[*] if (a == -1) {
[*] // 今天结束<span />
[*] set.clear();
[*] System.out.println("今天比赛结束1");
[*] stop--;
[*] continue;
[*] }
[*] set.add(a);
[*] setCheck.add(a);
[*] tempa = map.get(a);
[*] pointer++;
[*] if (pointer >= ids.length) {
[*] pointer = 0;
[*] }
[*] do {
[*] b = findNext(pointer);
[*] if (b == -1) {
[*] break;
[*] }
[*] setCheck.add(b);
[*] tempb = map.get(b);
[*] if (!tempa.contains(b) && !tempb.contains(a)) {
[*] break;
[*] }
[*] } while (b != -1);
[*] if (b == -1) {
[*] // 今天结束<span />
[*] set.clear();
[*] pointer--;
[*] System.out.println("今天比赛结束2");
[*] stop--;
[*] continue;
[*] }
[*] tempa.add(b);
[*] tempb.add(a);
[*] set.add(b);
[*] // pointer++;<span />
[*] System.out.println(a + "/" + b);
[*] total--;
[*] }
[*] }
[*]<span />
[*] static int findNext(int p) {
[*] int len = ids.length;
[*] while (len > 0) {
[*] if (p >= ids.length) {
[*] p = 0;
[*] }
[*] if (p < 0) {
[*] p = ids.length - 1;
[*] }
[*] if (!setCheck.contains(ids) && !set.contains(ids)) {
[*] return ids;
[*] }
[*] p++;
[*] len--;
[*] }
[*] return -1;
[*] }
[*]}
页:
[1]