将此图表作为参考,假设我想要0到5之间的最长路径.
那将是:0-> 1-> 3-> 2-> 4-> 6-> 5
对此有什么好的算法吗?我搜索过,但没有发现任何我能理解的内容.
我已经找到了用于最短路径的大量算法(0-> 1-> 2-> 4-> 6-> 5)并且我已经成功地实现了它们.
也许我是问题所在,但我想其他的想法:)
欢迎任何帮助
解决方法:
这个问题是NP-Hard(从哈密顿路径到你的问题有一个简单的减少,并且已知汉密尔顿路径搜索是NP难的).这意味着没有多项式解决这个问题(除非P = NP).
如果需要精确解,可以使用动态编程(指数状态):状态为(访问顶点的掩码,last_vertex),值为true或false.转换是添加一个新的顶点,如果在last_vertex和新顶点之间有一条边,则该顶点不在掩码中.它具有O(2 ^ n * n ^ 2)时间复杂度,仍然优于O(n!)回溯.
这是动态编程解决方案的伪代码:
f = array of (2 ^ n) * n size filled with false values
f(1 << start, start) = true
for mask = 0 ... (1 << n) - 1:
for last = 0 ... n - 1:
for new = 0 ... n - 1:
if there is an edge between last and new and mask & (1 << new) == 0:
f(mask | (1 << new), new) |= f(mask, last)
res = 0
for mask = 0 ... (1 << n) - 1:
if f(mask, end):
res = max(res, countBits(mask))
return res
还有一点关于从汉密尔顿路径到这个问题的减少:
def hamiltonianPathExists():
found = false
for i = 0 ... n - 1:
for j = 0 ... n - 1:
if i != j:
path = getLongestPath(i, j) // calls a function that solves this problem
if length(path) == n:
found = true
return found
这是一个Java实现(我没有正确测试,因此它可能包含错误):
/**
* Finds the longest path between two specified vertices in a specified graph.
* @param from The start vertex.
* @param to The end vertex.
* @param graph The graph represented as an adjacency matrix.
* @return The length of the longest path between from and to.
*/
public int getLongestPath(int from, int to, boolean[][] graph) {
int n = graph.length;
boolean[][] hasPath = new boolean[1 << n][n];
hasPath[1 << from][from] = true;
for (int mask = 0; mask < (1 << n); mask++)
for (int last = 0; last < n; last++)
for (int curr = 0; curr < n; curr++)
if (graph[last][curr] && (mask & (1 << curr)) == 0)
hasPath[mask | (1 << curr)][curr] |= hasPath[mask][last];
int result = 0;
for (int mask = 0; mask < (1 << n); mask++)
if (hasPath[mask][to])
result = Math.max(result, Integer.bitCount(mask));
return result;
}
标签:java,graph,data-structures
来源: https://codeday.me/bug/20190703/1361781.html