擅长:python、mysql、java
<p>我建议你建立一个二维的方阵。数组的维数应该是<code>N x N</code>。每个索引代表一个玩家。所以<code>passes[i][j]</code>处的值是玩家<code>i</code>传递给玩家<code>j</code>的次数。值<code>passes[i][i]</code>始终为零,因为玩家无法传递给自己</p>
<p>这里有一个例子。在</p>
<pre><code>players = ['Charles','Meow','Rebecca']
players = dict( zip(players,range(len(players)) ) )
rplayers = dict(zip(range(len(players)),players.keys()))
passes = []
for i in range(len(players)):
passes.append([ 0 for i in range(len(players))])
def pass_to(f,t):
passes[players[f]][players[t]] += 1
pass_to('Charles','Rebecca')
pass_to('Rebecca','Meow')
pass_to('Charles','Rebecca')
def showPasses():
for i in range(len(players)):
for j in range(len(players)):
print("%s passed to %s %d times" % ( rplayers[i],rplayers[j],passes[i][j],))
showPasses()
</code></pre>