Bridging signals

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 crossing each other, is imminent. Bearing in mind that there may be thousands 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? 

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 specifies 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

题意：由于芯片上电路焊错了，有些连线交叉，导致信号互相干扰。n组数据，每组有p个数据，存入数组a中；a[1]..a[i]..a[p]描述的是左边第i个端口和右边第a[i]个端口相连（上图）。现在可以桥接一些线路（剪掉？），求桥接后硅表面上互不交叉的线路的最大个数。

思路：从左边的端口来看这组数据的话，每个数据描述的是这个端口哪个端口相连。由于左边端口是递增的，那么最完美的情况就是 每个a[i]==i。因此想要求最优解就要求数列 a[]的最长上升子序列，因为 i 是递增的，所以对于上升子序列来说，i<i+1,a[i]<a[i+1]，因此这两条线路根本不会相交。可以根据上图模拟一下。

代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
#define inf 0x3f3f3f3f
#define N 40010
using namespace std;
int num[N],dp[N];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n;
        scanf("%d",&n);   
        for(int i=0;i<n;i++)  
            scanf("%d",&num[i]);
        memset(dp,inf,sizeof(dp));
        for(int i=0;i<n;i++)
            *lower_bound(dp,dp+n,num[i])=num[i]; 
        printf("%d\n",lower_bound(dp,dp+n,inf)-dp);
    }
    return 0;
}