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：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）。现在可以减掉一写线路，求减掉后硅表面上互不交叉的线路的最大个数。

思路：由于左边端口是递增的，如果每个a[i]==i的话这样肯定不会交叉了，但是这个线路是连接好的不能拆了重新连，所以就要求数列的最长上升子序列，如果i<i+1,a[i]<a[i+1]，那么这两条线路根本不会相交。需要注意的是这个题数据范围有点大，用l两层for循环会超时，所以用如下方法求最大上升子序列。代码如下：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define inf 0x3f3f3f3f
using namespace std;
int a[40010],dp[40010];
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int n;
        scanf("%d",&n);
        for(int i=0;i<n;i++)
            scanf("%d",&a[i]);
        memset(dp,inf,sizeof(dp));
        for(int i=0;i<n;i++)
            *lower_bound(dp,dp+n,a[i])=a[i];
        printf("%d\n",lower_bound(dp,dp+n,inf)-dp);
    }
    return 0;
}