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
64 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
題目本質：：求最長上升子序列（這裡沒有重復數字）。
我們有兩種思路求可以參考shuoj上的D序列的題目。這裡給出題目的題解鏈接：：
主要是兩種思路：：（1）lower_bound（2）二分法，如果覺得代碼不易理解可以點上面的鏈接
兩種方法的思路是一樣的。將數組A中子序列長度為 i 的最小值存放在數組S中。我們以3 2 4 6 5 7 3 為例進行演示行為遍歷，列為數組S，變化的地方已經標出來，有助於理解。在這裡a[ i ] > s[ j ]&&a[i]<=s[ j + 1 ]就應該把a[ i ]放在s[ j+1 ]的位置。所以關鍵就是找出 j 就知道把a[ i ]放在哪了。上面的兩種方法就是用來尋找 j的 。（在這裡lower_bound直接返回 j + 1 ） 我們可以發現s數組中的值必然是有序遞增的，這也是可以利用二分法的一個必要條件。
演示
0
1
2
3
4
1
3
2
2
3
2
4
4
2
4
6
5
2
4
5
6
2
4
5
7
7
2
3
5
7
這裡給出第二種方法代碼：：
#include
#include
#include
#define INF 0x3f3f3f3f
using namespace std;
const int N = 1e5 + 5;
int s[N];
int n,p,a[N];
int len;
int main()
{
cin>>n;
while(n--){
cin>>p;
memset(s,0,sizeof(s));
for(int i = 0;i>a[i];
s[1] = a[0];len = 1;//長度從1開始
for(int i = 1;is[len])s[++len] = a[i];
else{
/*************/int l = 1,r = len,mid;//這裡的二分法采用了左閉右閉的思路
int ans = 0;
while(l<=r)
{
mid = (l+r)/2;
if(s[mid]如果代碼不易理解請點擊鏈接，鏈接為：：
第一種的代碼只要將兩個/**************/之間的代碼換為
int p = lower_bound(s+1,s+len+1,t)-s;
s[p] = t;
就可以了。
3
9
1
4
3
9
1
4
兩種方法的思路是一樣的。將數組A中子序列長度為 i 的最小值存放在數組S中。我們以3 2 4 6 5 7 3 為例進行演示行為遍歷，列為數組S，變化的地方已經標出來，有助於理解。在這裡a[ i ] > s[ j ]&&a[i]<=s[ j + 1 ]就應該把a[ i ]放在s[ j+1 ]的位置。所以關鍵就是找出 j 就知道把a[ i ]放在哪了。上面的兩種方法就是用來尋找 j的 。（在這裡lower_bound直接返回 j + 1 ） 我們可以發現s數組中的值必然是有序遞增的，這也是可以利用二分法的一個必要條件。
演示
0
1
2
3
4
1
3
2
2
3
2
4
4
2
4
6
5
2
4
5
6
2
4
5
7
7
2
3
5
7
