Bridging signals

题目：

'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

Sample Output

题意：先输入一个数t，表示有t组数据，每组数据第一行有一个p，下面紧接着p行，1<= i <=p,

每行的数字a[i]表示右边的 i 和左边的a[i]相连接，找出在不相交的情况下有多少相连的；

思路：

动态规划的求最大上升子序列问题；

此题数组长度p<=40000, 所以没有优化的求法肯定超时，这里就用到了lower_bound（）函数；

用法：https://blog.csdn.net/yao166164474/article/details/53115372

就是定义dp[i]:=长度为i+1的上升子序列中末尾元素的最小值(不存在的话就是INF)，

初始化对dp[]数组全部赋值成INF，也就是+∞，然后对每个 aj 如果 i = 0 或者 dp[i-1] < aj 的话，

就用dp[i] = min(dp[i],aj)进行更新。最终找出使得dp[i]<INF的最大的i+1就是结果。

代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int a[40010],dp[40010];
int INF=0x3f3f3f3f;
int main()
{int t;scanf("%d",&t);while(t--){memset(dp,0x3f3f3f3f,sizeof(dp));memset(a,0,sizeof(a));int n;scanf("%d",&n);for(int i=1;i<=n;i++)scanf("%d",&a[i]);int maxx=0;for(int i=1;i<=n;i++){*lower_bound(dp+1,dp+n+1,a[i])=a[i];}printf("%d\n",lower_bound(dp+1,dp+n+1,INF)-dp-1);}return 0;
}

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