'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：新来的实习生把路由线路搞得一团糟！原本左右端口应当按顺序连接。现在只有切
除部分线路，使得任何线路都不相交。希望你写一个程序计算最后剩下多少线路？
思路：求上升公共子序列我们定义dp[i]:长度为i的上升子序列中末尾元素的最小值（不存在的话就是inf）
原理：如果子序列的长度相同，那么最末尾的元素较小的在之后会有优势。整个过程dp[i]是有序的，因此我们就可以采用二分查找来解决
代码：
#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define inf 0x3f3f3f
using namespace std;
int t,n;
int a[40000];
int dp[40000];
int main()
{
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
dp[i]=inf;//初始化为正无穷大
}
for(int i=1;i<=n;i++)
{
*lower_bound(dp+1,dp+1+n,a[i])=a[i];//lower_bound返回一个非递减序列[first, last)中的第一个大于等于值val的位置。因此我们加上*就是该位置所储存的变量
}
printf("%d\n",lower_bound(dp+1,dp+n+1,inf)-dp-1);//最后一个不为inf的数的下标即为答案
}
}