该代码的目标是:具有N个标记顶点和K个未标记边的简单连通图的数目。你知道吗
注意:这可能被认为是一个代码检查问题,但是经过反复尝试,我认为python和java代码具有相同的功能。我不确定代码是否有问题,或者与语言的复杂性有关,或者我忽略了某些东西的错误。
这是为了谷歌Foobar挑战赛。我用上面的方法完成了。我已经发布了链接到所有的源代码,测试所有可能的情况。你知道吗
第一种方法完全有效。唯一的问题-它进行O(NK)递归调用,K在N中是平均二次的[Full source]
一个朋友想出了一个算法,用自下而上的方法做同样的事情。主要功能:
def answerHelper(n,k):
totalGraphs = 0
for s in range(1,n):
graphs = 0
for t in range(0,k+1):
graphs += answer(s, t) * answer(n - s, k - t)
graphs = choose(n, s)*s*(n - s) * graphs
totalGraphs+= graphs
return totalGraphs/2
F = {}
def answer(n, k):
if (n, k) in F:
return F[n, k]
N = n * (n - 1)/2
if k is n - 1:
return int(n ** (n-2))
if k < n or k > N:
return 0
if k == N:
return 1
result = ((N - k + 1) * answer(n, k - 1) + answerHelper(n, k - 1)) / k
F[n, k] = result
return result
与原始工作Java代码 [diffchecker]相比,python在4种情况下失败。我想这是因为某种溢出(?)。[Full python source]
我正在尝试将这个python代码转换成Java。这就是我想到的。你知道吗
static Map<List<Integer>, String> resultMap = new HashMap<List<Integer>, String>();
public static String answer(int N, int K) {
/* for the case where K > N-1 */
// check if key is present in the map
List<Integer> tuple = Arrays.asList(N, K);
if( resultMap.containsKey(tuple) )
return resultMap.get(tuple);
// maximum number of edges in a simply
// connected undirected unweighted graph
// with n nodes = |N| * |N-1| / 2
int maxEdges = N * (N-1) / 2;
/* for the case where K < N-1 or K > N(N-1)/2 */
if(K < N-1 || K > maxEdges)
return BigInteger.ZERO.toString();
/* for the case where K = N-1 */
// Cayley's formula applies [https://en.wikipedia.org/wiki/Cayley's_formula].
// number of trees on n labeled vertices is n^{n-2}.
if(K == N-1)
return BigInteger.valueOf((long)Math.pow(N, N-2)).toString();
/* for the case where K = N(N-1)/2 */
// if K is the maximum possible
// number of edges for the number of
// nodes, then there is only one way is
// to make a graph (connect each node
// to all other nodes)
if(K == maxEdges)
return BigInteger.ONE.toString();
// number of edges left from maxEdges if I take away K-1 edges
BigInteger numWays = BigInteger.valueOf(maxEdges - K + 1);
// number of graphs possible for each of the numWays edges for a graph that has 1 less edge
BigInteger numGraphsWithOneLessEdge = new BigInteger( answer(N, K-1) );
// number of all possible subgraphs with K-1 edges
BigInteger subGraphs = answerHelper(N, K-1);
// numWays*numGraphsWithOneLessEdge + subGraphs
BigInteger result = subGraphs.add(numWays.multiply(numGraphsWithOneLessEdge));
// this contains repeats for each of the K edges
result = result.divide(BigInteger.valueOf(K));
// add to cache
resultMap.put(Collections.unmodifiableList(Arrays.asList(N, K)), result.toString());
return resultMap.get(tuple);
}
private static BigInteger answerHelper(int N, int K) {
BigInteger totalGraphs = BigInteger.ZERO;
for(int n = 1 ; n < N ; n++) {
BigInteger graphs = BigInteger.ZERO;
for(int k = 0 ; k <= K ; k++) {
// number of graphs with n nodes and k edges
BigInteger num = new BigInteger( answer(n, k) );
// number of graphs with N-n nodes and K-k edges
BigInteger num2 = new BigInteger( answer(N-n, K-k) );
graphs = graphs.add( num.multiply(num2) );
}
// number of ways to choose n nodes from N nodes
BigInteger choose = choose(N, n);
// this is repeated for each of the n chosen nodes
// and the N-n unchosen nodes
choose = choose.multiply(BigInteger.valueOf(n)).multiply(BigInteger.valueOf(N-n));
totalGraphs = totalGraphs.add( choose.multiply(graphs) );
}
// now, consider the case where N = 20
// when n = 2, we compute for N-n = 18
// when n = 18, we do the same thing again
// hence, contents of totalGraphs is 2 times
// of what it should be
return totalGraphs.divide(BigInteger.valueOf(2));
}
这段代码的功能与Python相同,但与工作的java代码[diffchecker]相比,它在多种情况下都失败了
如果我能得到一些指导,我将非常感激。你知道吗
问题出在Java代码中,而不是Python代码中(我怀疑是溢出;一些细致的调试证明不是这样)。这不是最简单的比较数字与20个奇数)。你知道吗
错误代码:
对于N>;=17,
(long)Math.pow(N, N-2)
不准确。这是因为双值越大,连续值之间的差距就越大。double不能表示其范围内的所有整数值,这就是这里的问题所在。它返回最接近精确结果的双精度值。此外,对于双值,尾数是52位,大约等于16(?)小数位数。因此,泛滥(其实不是一个词)。 因此,返回的值比应该的值小。必须用下面的代码块替换它。你知道吗相关问题 更多 >
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