poj1631(最长上升子序列 nlogn)

题目见http://poj.org/problem?id=1631

最长上升子序列（LIS）裸题（这么说相信题目看看也懂了吧）

多数据，由于每个数据p<40000，最多可以有上万个数，这个时候用O(n^2)算法显然就不可行了

然后就要利用O(nlogn)算法去做

点击打开链接 这个博客讲的还是不错的，比较好懂

然后不多说，照样子写个模板实现实现，代码如下

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

class Reader {
	static BufferedReader reader;
	static StringTokenizer tokenizer;

	static void init(InputStream input) {
		reader = new BufferedReader(new InputStreamReader(input));
		tokenizer = new StringTokenizer("");
	}

	static String next() throws IOException {
		while (!tokenizer.hasMoreTokens()) {
			tokenizer = new StringTokenizer(reader.readLine());
		}
		return tokenizer.nextToken();
	}

	static int nextInt() throws IOException {
		return Integer.parseInt(next());
	}
}

public class Main {

	/**
	 * @param args
	 */
	static int t, n, len, l, r, k, mid;
	static int arr[], g[];

	private static int find(int num) {
		l = 0;
		r = len;
		while (r - l > 1) {
			mid = (l + r) / 2;
			if (g[mid] >= num)
				r = mid;
			else
				l = mid;
		}
		return l;
	}

	public static void main(String[] args) throws IOException {
		// TODO Auto-generated method stub
		Reader.init(System.in);
		t = Reader.nextInt();
		for (int casenum = 1; casenum <= t; casenum++) {
			n = Reader.nextInt();
			arr = new int[n + 1];
			for (int i = 1; i <= n; i++)
				arr[i] = Reader.nextInt();
			g = new int[n + 1];
			g[0] = 0;
			len = 0;
			for (int i = 1; i <= n; i++) {
				if (arr[i] > g[len]) {
					len++;
					g[len] = arr[i];
				} else {
					k = find(arr[i]);
					g[k + 1] = arr[i];
				}
			}
			System.out.println(len);
		}
	}

}