解题思路：题目给出的描述就是一种求最长上升子序列的方法 将该列数an与其按升序排好序后的an'求出最长公共子序列就是最长上升子序列
但是这道题用这种方法是会超时的，用滚动数组优化也超时，
下面是网上找的求LIS的算法
假设要寻找最长上升子序列的序列是a[n]，然后寻找到的递增子序列放入到数组b中。
（1）当遍历到数组a的第一个元素的时候，就将这个元素放入到b数组中，以后遍历到的元素都和已经放入到b数组中的元素进行比较；
（2）如果比b数组中的每个元素都大，则将该元素插入到b数组的最后一个元素，并且b数组的长度要加1；
（3）如果比b数组中最后一个元素小，就要运用二分法进行查找，查找出第一个比该元素大的最小的元素，然后将其替换。
在这个过程中，只重复执行这两步就可以了，最后b数组的长度就是最长的上升子序列长度。
例如样例给的 4 2 6 3 1 5
那么
4//
2//用2替换4
2 6//加入6
2 3//用3替换6
1 3//用1替换2
1 3 5//加入5
注意最后得到的a数组中的数为1 3 5
而实际上我们要求的最长上升序列为 2 3 5 但是我们只需要求长度，所以不影响
Bridging signals
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 736 Accepted Submission(s): 486
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.
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
#include<stdio.h>
int b[500010],f[500010];
int bserch(int f[],int n,int v)
{
int x,y,m;
x=1;
y=n;
while(x<=y)
{
m=(x+y)/2;
if(v>=f[m])
x=m+1;
else
y=m-1;
}
return x;
}
int main()
{
int i, n,len,ncase,tmp,x,y;
scanf("%d",&ncase);
while(ncase--)
{
scanf("%d",&n);
for(i=1;i<=n;i++)
scanf("%d",&b[i]);
len=1;
f[1]=b[1];
for(i=2;i<=n;i++)
{
if(b[i]>=f[len])
{
len++;
f[len]=b[i];
}
else
{
tmp=bserch(f,len,b[i]);
f[tmp]=b[i];
}
}
printf("%d
",len);
}
}