poj1631(最长上升子序列 nlogn)

本文介绍了一种高效算法,用于解决信号交叉问题,通过寻找序列中的最长上升子序列来最小化信号桥接次数。使用二分查找优化,避免了暴力解决的超时问题,特别适用于大规模信号路由设计。

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'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:

第一个数据t代表t组数据,然后每组数据有一个n,紧接着有n个数字,分别是1-n的错乱排序,然后求这个序列的最长上升子序列。

思路:

用暴力解决超时,于是在查找插入的时候采用2分法,轻松解决问题,但是poj提交测试数据有问题,用c++输入输出流会超时,用scanf,printf可以通过。

#include<iostream>
#include <cstdio>
using namespace std;
int dp[40005];
int main()
{
    int i,T,n,top;
    int m;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        top=0;
        dp[top]=-1;
        for (i=1; i<=n; i++)
        {
            scanf("%d",&m);
            if (m>dp[top])
                {
                    top++;
                    dp[top]=m;

                }
            else
            {
                int low=1;
                int high=top;
                while(low<=high)
                {
                    int middle=(low+high)/2;
                    if (dp[middle]<m)
                        low=middle+1;
                    else
                        high=middle-1;
                }
                dp[low]=m;
            }

        }
        printf("%d\n",top);
    }
    return 0;
}
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