Bridging signals（DP）

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 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

两行1到n的数，第一行每个数对应第二行一个数，一一对应关系，从中最多能选出多少互不相交的线段。

这题没看出来- -，觉得竖着看太有误导性了，然后横着画了一个样例，看了一下，我操，这不是最大上升子序列吗- -，裸地。

直接O(nlogn)上就行了。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<queue>
#include<vector>
#include<algorithm>
#include<string>
#include<cmath>
#include<set>
#include<map>
#include<vector>
#include<stack>
#include<utility>
#include<sstream>
using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int maxn = 1005;
int mp[40005],dp[40005];
int main()
{
    #ifdef LOCAL
    freopen("C:\\Users\\ΡΡ\\Desktop\\in.txt","r",stdin);
    //freopen("C:\\Users\\ΡΡ\\Desktop\\out.txt","w",stdout);
    #endif // LOCAL
    int n;
    scanf("%d",&n);
    while(n--)
    {
        int p;
        scanf("%d",&p);
        for(int i = 1;i <= p;i++)
            scanf("%d",&mp[i]);
        for(int i = 0;i <= p;i++)
            dp[i] = inf;
        for(int i = 1;i <= p;i++)
        {
            int x = lower_bound(dp + 1,dp + 1 + p,mp[i]) - dp;
            dp[x] = min(dp[x],mp[i]);
        }
        int ans = lower_bound(dp + 1,dp + 1 + p,inf) - 1 - dp;
        printf("%d\n",ans);
    }
    return 0;
}