poj1631(Bridging signals)

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

其实就是求最长上升子序列(LIS)，由于长度太长，所以要用优化过的算法，复杂度降到了O(nlogn)，dp[i]表示长度为i的LIS尾元素的最小值，当前获得的最大长度为len。若num[i]>dp[len]，则dp[++len]=num[i]，否则的话，注意到dp数组肯定是一个单调递增的数组，所以要找到一个j，使得dp[j]<num[i]<dp[j+1]，再令dp[j+1]=num[i]，查找可以使用二分，c++提供了upper_bound函数。这样最后的len就是所求答案。

#include <iostream>
#include <sstream>
#include <fstream>
#include <string>
#include <map>
#include <vector>
#include <list>
#include <set>
#include <stack>
#include <queue>
#include <deque>
#include <algorithm>
#include <functional>
#include <iomanip>
#include <limits>
#include <new>
#include <utility>
#include <iterator>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cmath>
#include <ctime>
using namespace std;
typedef long long LL;
const int INF = 0x3f3f3f3f;
const double PI = acos(-1.0);
const double EPS = 1e-8;
const int MAXN = 1000010;
int dx[] = {0, 1, 0, -1}, dy[] = {-1, 0, 1, 0};

int main()
{
    int T;
    cin >> T;
    while (T--)
    {
        int n, num, d[40010], len = 0;
        cin >> n;
        memset(d, 0, sizeof(d));
        for (int i = 0; i < n; ++i)
        {
            scanf("%d", &num);
            if (num > d[len])
                d[++len] = num;
            else
                *upper_bound(d, d+len, num) = num;
        }
        cout << len << endl;
    }
    return 0;
}