求序列的最长递增子序列

[bazookier]

请给出一个O(nlgn)时间的算法，使之能找出一个n个数的序列中最长的单调递增子序列。(提示：观察长度为i的一个候选子序列的最后一个元素，它至少与长度为i-1的一个候选子序列的最后一个元素一样大。通过把候选子序列与输入序列相连接来维护它们)。

[zjk2752]

不明白你那提示是什么意思

[wuyi8808]

比较麻烦，看看算法的书。

[Yoon_EunHae]

是公共递增子序列？ 这个最快的算法是nlogn的 2分搜索
还是某个串递增子序列？ 这个可以线性规划的0（n）
经典的dp题,算法到处都是..！

[可口可乐]

参考网页
http://www.algorithmist.com/index.php/Longest_Increasing_Subsequence

Faster Algorithm

There's also an O(nlogn) solution based on some observations. Let Ai,j be the smallest possible tail out of all increasing subsequences of length j using elements a_1, a_2, a_3, \ldots, a_i.

Observe that, for any particular i, A_{i,1} < A_{i,2} < \ldots < A_{i,j}. This suggests that if we want the longest subsequence that ends with ai + 1, we only need to look for a j such that Ai,j < ai + 1 < = Ai,j + 1 and the length will be j + 1.

Notice that in this case, Ai + 1,j + 1 will be equal to ai + 1, and all Ai + 1,k will be equal to Ai,k for k \ne j + 1.

Furthermore, there is at most one difference between the set Ai and the set Ai + 1, which is caused by this search.

Since A is always ordered in increasing order, and the operation does not change this ordering, we can do a binary search for every single a_1, a_2, \ldots, a_n.

#include <vector>
using namespace std;
 
/* Finds longest strictly increasing subsequence. O(n log k) algorithm. */
template<typename T> vector<int> find_lis(vector<T> &a)
{
	vector<int> b, p(a.size());
	int u, v;
 
	if (a.size() < 1) return b;
 
	b.push_back(0);
 
	for (int i = 1; i < (int)a.size(); i++) {
		if (a[b.back()] < a[i]) {
			p[i] = b.back();
			b.push_back(i);
			continue;
		}
 
		for (u = 0, v = b.size()-1; u < v;) {
			int c = (u + v) / 2;
			if (a[b[c]] < a[i]) u=c+1; else v=c;
		}
 
		if (a[i] < a[b[u]]) {
			if (u > 0) p[i] = b[u-1];
			b[u] = i;
		}	
	}
 
	for (u = b.size(), v = b.back(); u--; v = p[v]) b[u] = v;
	return b;
}
 
/* Example of usage: */
#include <cstdio>
int main()
{
	int a[] = { 1, 9, 3, 8, 11, 4, 5, 6, 4, 19, 7, 1, 7 };
	vector<int> seq(a, a+sizeof(a)/sizeof(a[0]));
	vector<int> lis = find_lis(seq);
 
	for (unsigned i = 0; i < lis.size(); i++)
		printf(i+1 < lis.size() ? "%d " : "%d\n", seq[lis[i]]);
 
	return 0;
}

[bazookier]

看不懂，知道是对的但不知道为什么。英文解释太抽象了，没看明白。希望给解释下。

[大王派我去巡山]

啥也不说了，直接上图：

[michaelwangwh]

动态规划,10年前就碰到这样的题目了,拦截导弹,太经典了

[dzrjojo]

good