最长上升子序列nlogn算法

Link:http://acm.hdu.edu.cn/showproblem.php?pid=1950

Bridging signals

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 716 Accepted Submission(s): 475



Problem Description

'Oh no, they've done it again', cries the chief designer at the Waferland chip factory. Once more the routing designers have screwed up completely, making the signals on the chip connecting the ports of two functional blocks cross each other all over the place.
At this late stage of the process, it is too

expensive to redo the routing. Instead, the engineers have to bridge the signals, using the third dimension, so that no two signals cross. However, bridging is a complicated operation, and thus it is desirable to bridge as few signals as possible. The call
for a computer program that finds the maximum number of signals which may be connected on the silicon surface without rossing each other, is imminent. Bearing in mind that there may be housands of signal ports at the boundary of a functional block, the problem
asks quite a lot of the programmer. Are you up to the task?



Figure 1. To the left: The two blocks' ports and their signal mapping (4,2,6,3,1,5). To the right: At most three signals may be routed on the silicon surface without crossing each other. The dashed signals must be bridged.

A typical situation is schematically depicted in figure 1. The ports of the two functional blocks are numbered from 1 to p, from top to bottom. The signal mapping is described by a permutation of the numbers 1 to p in the form of a list of p unique numbers
in the range 1 to p, in which the i:th number pecifies which port on the right side should be connected to the i:th port on the left side.

Two signals cross if and only if the straight lines connecting the two ports of each pair do.



Input

On the first line of the input, there is a single positive integer n, telling the number of test scenarios to follow. Each test scenario begins with a line containing a single positive integer p<40000, the number of ports on the two functional blocks. Then
follow p lines, describing the signal mapping: On the i:th line is the port number of the block on the right side which should be connected to the i:th port of the block on the left side.



Output

For each test scenario, output one line containing the maximum number of signals which may be routed on the silicon surface without crossing each other.



Sample Input

4
6
4
2
6
3
1
5
10
2
3
4
5
6
7
8
9
10
1
8
8
7
6
5
4
3
2
1
9
5
8
9
2
3
1
7
4
6

Sample Output

3
9
1
4

Source

NWERC2003

hdu 1950 Bridging signals

http://acm.hdu.edu.cn/showproblem.php?pid=1950

===================================

最长上升子序列（LIS）的典型变形，熟悉的n^2的动归会超时。LIS问题可以优化为nlogn的算法。

定义d[k]:长度为k的上升子序列的最末元素，若有多个长度为k的上升子序列，则记录最小的那个最末元素。

注意d中元素是单调递增的，下面要用到这个性质。

首先len = 1,d[1] = a[1],然后对a[i]:若a[i]>d[len]，那么len++，d[len] = a[i];

否则，我们要从d[1]到d[len-1]中找到一个j，满足d[j-1]<a[i]<d[j],则根据D的定义，我们需要更新长度为j的上升子序列的最末元素（使之为最小的）即 d[j] = a[i];

最终答案就是len

利用d的单调性，在查找j的时候可以二分查找，从而时间复杂度为nlogn。

==================================

最长上升子序列nlogn算法

在川大oj上遇到一道题无法用n^2过于是，各种纠结，最后习得nlogn的算法

最长递增子序列，Longest Increasing Subsequence 下面我们简记为 LIS。

排序+LCS算法 以及 DP算法就忽略了，这两个太容易理解了。

假设存在一个序列d[1..9] = 2 1 5 3 6 4 8 9 7，可以看出来它的LIS长度为5。n

下面一步一步试着找出它。

我们定义一个序列B，然后令 i = 1 to 9 逐个考察这个序列。

此外，我们用一个变量Len来记录现在最长算到多少了

首先，把d[1]有序地放到B里，令B[1] = 2，就是说当只有1一个数字2的时候，长度为1的LIS的最小末尾是2。这时Len=1

然后，把d[2]有序地放到B里，令B[1] = 1，就是说长度为1的LIS的最小末尾是1，d[1]=2已经没用了，很容易理解吧。这时Len=1

接着，d[3] = 5，d[3]>B[1]，所以令B[1+1]=B[2]=d[3]=5，就是说长度为2的LIS的最小末尾是5，很容易理解吧。这时候B[1..2] = 1, 5，Len＝2

再来，d[4] = 3，它正好加在1,5之间，放在1的位置显然不合适，因为1小于3，长度为1的LIS最小末尾应该是1，这样很容易推知，长度为2的LIS最小末尾是3，于是可以把5淘汰掉，这时候B[1..2] = 1, 3，Len = 2

继续，d[5] = 6，它在3后面，因为B[2] = 3, 而6在3后面，于是很容易可以推知B[3] = 6, 这时B[1..3] = 1, 3, 6，还是很容易理解吧？ Len = 3 了噢。

第6个, d[6] = 4，你看它在3和6之间，于是我们就可以把6替换掉，得到B[3] = 4。B[1..3] = 1, 3, 4， Len继续等于3

第7个, d[7] = 8，它很大，比4大，嗯。于是B[4] = 8。Len变成4了

第8个, d[8] = 9，得到B[5] = 9，嗯。Len继续增大，到5了。

最后一个, d[9] = 7，它在B[3] = 4和B[4] = 8之间，所以我们知道，最新的B[4] =7，B[1..5] = 1, 3, 4, 7, 9，Len = 5。

于是我们知道了LIS的长度为5。

!!!!! 注意。这个1,3,4,7,9不是LIS，它只是存储的对应长度LIS的最小末尾。有了这个末尾，我们就可以一个一个地插入数据。虽然最后一个d[9] = 7更新进去对于这组数据没有什么意义，但是如果后面再出现两个数字 8 和 9，那么就可以把8更新到d[5], 9更新到d[6]，得出LIS的长度为6。

然后应该发现一件事情了：在B中插入数据是有序的，而且是进行替换而不需要挪动——也就是说，我们可以使用二分查找，将每一个数字的插入时间优化到O(logN)~~~~~于是算法的时间复杂度就降低到了O(NlogN)～！

/*
	HDU 1950 Bridging signals
			-----最长上升子序列nlogn算法
*/

#include<cstdio>
#include<cstring>
#define MAXN 40005

int arr[MAXN],ans[MAXN],len;

/* 
	二分查找。 注意，这个二分查找是求下界的;  (什么是下界？详情见《算法入门经典》 P145)
	即返回 >= 所查找对象的第一个位置（想想为什么）

	也可以用STL的lowe_bound二分查找求的下界
*/

int binary_search(int i){
	int left,right,mid;
	left=0,right=len;
	while(left<right){
		mid = left+(right-left)/2;
		if(ans[mid]>=arr[i]) right=mid;
		else left=mid+1;
	}
	return left;
}

int main()
{
  	//freopen("input.txt","r",stdin);
	int T,p,i,j,k;
	scanf("%d",&T);
	while(T--){
		scanf("%d",&p);
		for(i=1; i<=p; ++i)
			scanf("%d",&arr[i]);
		
		ans[1] = arr[1];
		len=1;
		for(i=2; i<=p; ++i){
			if(arr[i]>ans[len])
				ans[++len]=arr[i];
			else{
				int pos=binary_search(i);   // 如果用STL： pos=lower_bound(ans,ans+len,arr[i])-ans; 
				ans[pos] = arr[i];
		}
	}
		printf("%d\n",len);
	}
	return 0;
}