Baseball Elimination - Princeton Part II Programming Assignment 3

http://coursera.cs.princeton.edu/algs4/assignments/baseball.html
The baseball elimination problem.   In the baseball elimination problem, there is a division consisting of N teams. At some point during the season, team i has w[i] wins, l[i] losses, r[i] remaining games, and g[i][j]games left to play against team j. A team is mathematically eliminated if it cannot possibly finish the season in (or tied for) first place. The goal is to determine exactly which teams are mathematically eliminated. For simplicity, we assume that no games end in a tie (as is the case in Major League Baseball) and that there are no rainouts (i.e., every scheduled game is played).
The problem is not as easy as many sports writers would have you believe, in part because the answer depends not only on the number of games won and left to play, but also on the schedule of remaining games. To see the complication, consider the following scenario:

 
                w[i] l[i] r[i]        g[i][j]
i  team         wins loss left   Atl Phi NY  Mon
------------------------------------------------
0  Atlanta       83   71    8     -   1   6   1
1  Philadelphia  80   79    3     1   -   0   2
2  New York      78   78    6     6   0   -   0
3  Montreal      77   82    3     1   2   0   -

Montreal is mathematically eliminated since it can finish with at most 80 wins and Atlanta already has 83 wins. This is the simplest reason for elimination. However, there can be more complicated reasons. For example, Philadelphia is also mathematically eliminated. It can finish the season with as many as 83 wins, which appears to be enough to tie Atlanta. But this would require Atlanta to lose all of its remaining games, including the 6 against New York, in which case New York would finish with 84 wins. We note that New York is not yet mathematically eliminated despite the fact that it has fewer wins than Philadelphia.
It is sometimes not so easy for a sports writer to explain why a particular team is mathematically eliminated. Consider the following scenario from the American League East on August 30, 1996:

                w[i] l[i] r[i]          g[i][j]
i  team         wins loss left   NY Bal Bos Tor Det
---------------------------------------------------
0  New York      75   59   28     -   3   8   7   3
1  Baltimore     71   63   28     3   -   2   7   7
2  Boston        69   66   27     8   2   -   0   3
3  Toronto       63   72   27     7   7   0   -   3
4  Detroit       49   86   27     3   7   3   3   -

It might appear that Detroit has a remote chance of catching New York and winning the division because Detroit can finish with as many as 76 wins if they go on a 27-game winning steak, which is one more than New York would have if they go on a 28-game losing streak. Try to convince yourself that Detroit is already mathematically eliminated. Here's one ad hoc explanation; we will present a simpler explanation below.
A maxflow formulation.   We now solve the baseball elimination problem by reducing it to the maxflow problem. To check whether team x is eliminated, we consider two cases.

• Trivial elimination. If the maximum number of games team x can win is less than the number of wins of some other team i, then team x is trivially eliminated (as is Montreal in the example above). That is, if w[x] +r[x] < w[i], then team x is mathematically eliminated.
• Nontrivial elimination. Otherwise, we create a flow network and solve a maxflow problem in it. In the network, feasible integral flows correspond to outcomes of the remaining schedule. There are vertices corresponding to teams (other than team x) and to remaining divisional games (not involving team x). Intuitively, each unit of flow in the network corresponds to a remaining game. As it flows through the network from s to t, it passes from a game vertex, say between teams i and j, then through one of the team vertices i or j, classifying this game as being won by that team.More precisely, the flow network includes the following edges and capacities.
  
  
  
  • We connect an artificial source vertex s to each game vertex i-j and set its capacity to g[i][j]. If a flow uses all g[i][j] units of capacity on this edge, then we interpret this as playing all of these games, with the wins distributed between the team vertices i and j.
  • We connect each game vertex i-j with the two opposing team vertices to ensure that one of the two teams earns a win. We do not need to restrict the amount of flow on such edges.
  • Finally, we connect each team vertex to an artificial sink vertex t. We want to know if there is some way of completing all the games so that team x ends up winning at least as many games as team i. Since teamx can win as many as w[x] + r[x] games, we prevent team i from winning more than that many games in total, by including an edge from team vertex i to the sink vertex with capacity w[x] + r[x] - w[i].
  If all edges in the maxflow that are pointing from s are full, then this corresponds to assigning winners to all of the remaining games in such a way that no team wins more games than x. If some edges pointing from sare not full, then there is no scenario in which team x can win the division. In the flow network below Detroit is team x = 4.
  What the min cut tells us.   By solving a maxflow problem, we can determine whether a given team is mathematically eliminated. We would also like to explain the reason for the team's elimination to a friend in nontechnical terms (using only grade-school arithmetic). Here's such an explanation for Detroit's elimination in the American League East example above. With the best possible luck, Detroit finishes the season with 49 + 27 = 76 wins. Consider the subset of teams R = { New York, Baltimore, Boston, Toronto }. Collectively, they already have 75 + 71 + 69 + 63 = 278 wins; there are also 3 + 8 + 7 + 2 + 7 = 27 remaining games among them, so these four teams must win at least an additional 27 games. Thus, on average, the teams in R win at least 305 / 4 = 76.25 games. Regardless of the outcome, one team in R will win at least 77 games, thereby eliminating Detroit.

In fact, when a team is mathematically eliminated there always exists such a convincing certificate of elimination, where R is some subset of the other teams in the division. Moreover, you can always find such a subset Rby choosing the team vertices on the source side of a min s-t cut in the baseball elimination network. Note that although we solved a maxflow/mincut problem to find the subset R, once we have it, the argument for a team's elimination involves only grade-school algebra.
Your assignment.   Write an immutable data type BaseballElimination that represents a sports division and determines which teams are mathematically eliminated by implementing the following API:

public BaseballElimination(String filename)                    // create a baseball division from given filename in format specified below
public              int numberOfTeams()                        // number of teams
public Iterable<String> teams()                                // all teams
public              int wins(String team)                      // number of wins for given team
public              int losses(String team)                    // number of losses for given team
public              int remaining(String team)                 // number of remaining games for given team
public              int against(String team1, String team2)    // number of remaining games between team1 and team2
public          boolean isEliminated(String team)              // is given team eliminated?
public Iterable<String> certificateOfElimination(String team)  // subset R of teams that eliminates given team; null if not eliminated

https://github.com/zhichaoh/Coursera-Algorithms/blob/master/src/BaseballElimination.java
public class BaseballElimination {

private int[] w;

private int[] l;

private int[] r;

private int[][] g;

private Map<String, Integer> teamNames;

private String[] nameList;

private int numTeams;

public BaseballElimination(String filename) throws Exception {
// create a baseball division from given filename in format specified
// below
Scanner input = new Scanner(new File(filename));
numTeams = input.nextInt();
this.w = new int[numTeams];
this.l = new int[numTeams];
this.r = new int[numTeams];
this.g = new int[numTeams][numTeams];
this.nameList = new String[numTeams];
this.teamNames = new HashMap<String, Integer>();
for (int n = 0; n < numTeams; n++) {
String name = input.next();
teamNames.put(name, n);
w[n] = input.nextInt();
l[n] = input.nextInt();
r[n] = input.nextInt();
for (int m = 0; m < numTeams; m++)
g[n][m] = input.nextInt();
nameList[n] = name;
}
input.close();
}

public int numberOfTeams() {
return this.numTeams;
}

public Iterable<String> teams() {
// all teams
return teamNames.keySet();
}

public int wins(String team) {
// number of wins for given team
if (!teamNames.containsKey(team))
throw new java.lang.IllegalArgumentException();
return w[teamNames.get(team)];
}

public int losses(String team) {
// number of losses for given team
if (!teamNames.containsKey(team))
throw new java.lang.IllegalArgumentException();
return l[teamNames.get(team)];
}

public int remaining(String team) {
// number of remaining games for given team
if (!teamNames.containsKey(team))
throw new java.lang.IllegalArgumentException();
return r[teamNames.get(team)];
}

public int against(String team1, String team2) {
// number of remaining games between team1 and team2
if (!teamNames.containsKey(team1) || !teamNames.containsKey(team2))
throw new java.lang.IllegalArgumentException();
int a = teamNames.get(team1);
int b = teamNames.get(team2);
return g[a][b];
}

public boolean isEliminated(String team) {
// is given team eliminated?
if (!teamNames.containsKey(team))
throw new java.lang.IllegalArgumentException();
int x = teamNames.get(team);

for (int i = 0; i < this.numTeams; i++) {
if (w[x] + r[x] - w[i] < 0)
return true;
}

int numMatches = this.numTeams * (this.numTeams - 1) / 2;
int nodeID = 0;

FlowNetwork fn = new FlowNetwork(numMatches + numTeams + 2);
int s = numMatches + numTeams;
int t = s + 1;

for (int i = 0; i < numTeams; i++) {
for (int j = i + 1; j < numTeams; j++) {
if (i == j)
continue;
fn.addEdge(new FlowEdge(s, nodeID, g[i][j])); // source to match
// nodes
fn.addEdge(new FlowEdge(nodeID, numMatches + i,
Integer.MAX_VALUE)); // match to team nodes
fn.addEdge(new FlowEdge(nodeID, numMatches + j,
Integer.MAX_VALUE)); // match to team nodes
nodeID += 1;
}
fn.addEdge(new FlowEdge(numMatches + i, t, Math.max(0, w[x] + r[x]
- w[i]))); // game nodes to target
}

new FordFulkerson(fn, s, t);

for (FlowEdge e : fn.adj(s)) {
if (e.flow() != e.capacity())
return true;
}
return false;
}

public Iterable<String> certificateOfElimination(String team) {
// subset R of teams that eliminates given team; null if not eliminated
if (!teamNames.containsKey(team))
throw new java.lang.IllegalArgumentException();
int x = teamNames.get(team);
int numMatches = this.numTeams * (this.numTeams - 1) / 2;
int nodeID = 0;

List<String> nList = new ArrayList<String>();
for (int i = 0; i < this.numTeams; i++) {
if (w[x] + r[x] - w[i] < 0)
nList.add(nameList[i]);
}
if (nList.size() > 0)
return nList;

FlowNetwork fn = new FlowNetwork(numMatches + numTeams + 2);
int s = numMatches + numTeams;
int t = s + 1;

for (int i = 0; i < numTeams; i++) {
for (int j = i + 1; j < numTeams; j++) {
if (i == j)
continue;
fn.addEdge(new FlowEdge(s, nodeID, g[i][j])); // source to match
// nodes
fn.addEdge(new FlowEdge(nodeID, numMatches + i,
Integer.MAX_VALUE)); // match to team nodes
fn.addEdge(new FlowEdge(nodeID, numMatches + j,
Integer.MAX_VALUE)); // match to team nodes
nodeID += 1;
}
fn.addEdge(new FlowEdge(numMatches + i, t, Math.max(0, w[x] + r[x]
- w[i]))); // game nodes to target
}

FordFulkerson FF = new FordFulkerson(fn, s, t);

boolean flag = false;
for (FlowEdge e : fn.adj(s)) {
if (e.flow() != e.capacity()) {
flag = true;
break;
}
}
if (!flag)
return null;
else {
List<Integer> nodeList = this.BFSRes(fn, s);
List<String> nl = new ArrayList<String>();
for (Integer v : nodeList) {
if (FF.inCut(v) && v >= numMatches) {
nl.add(this.nameList[v - numMatches]);
}
}
return nl;
}
}

private List<Integer> BFSRes(FlowNetwork graph, int node) {
Queue<Integer> Q = new Queue<Integer>();
boolean[] visited = new boolean[graph.V()];
Q.enqueue(node);
visited[node] = true;
List<Integer> nodeList = new ArrayList<Integer>();
while (!Q.isEmpty()) {
int cn = Q.dequeue();
for (FlowEdge e : graph.adj(cn)) {
int t = -1;
if (e.from() == cn)
t = e.to();
else
t = e.from();
if (e.residualCapacityTo(t) > 0) {
if (!visited[t]) {
Q.enqueue(t);
visited[t] = true;
nodeList.add(t);
}
}
}
}
return nodeList;
}

public static void main(String[] args) throws Exception {
BaseballElimination division = new BaseballElimination(args[0]);
for (String team : division.teams()) {
if (division.isEliminated(team)) {
StdOut.print(team + " is eliminated by the subset R = { ");
for (String t : division.certificateOfElimination(team))
StdOut.print(t + " ");
StdOut.println("}");
} else {
StdOut.println(team + " is not eliminated");
}
}
}
}
https://segmentfault.com/a/1190000005345079

本次作业在细节上有一些非常简单的小技巧可以帮助简化实现（简洁性也是checklist提示的，参考代码不到200行，我的版本为150行左右），但可能不是第一次思考时就能想到：

• 在读取文件数据时就可额外计算（保存）一些后期需要的数据：目前榜首队伍与获胜场数（用于简单淘汰），和以双方队伍为单位的比赛场次（用于建立FlowNetwork的顶点数）；
• "Do not worry about over-optimizing your program because the data sets that arise in real applications are tiny." 对我的思路，这意味着无需考虑FlowNetwork的复用问题——一个早期造成很多痛苦的优化尝试；
• FordFulkerson类的使用实际上使问题仅剩下了建模部分，且在说明中已解释得非常详细，故实现十分直接。