最长递增子序列(LIS)：O(nlogn)

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C - LIS(NlgN)

 

'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 rossing each other, is imminent. Bearing in mind that there may be housands 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?

Figure 1. To the left: The two blocks' ports and their signal mapping (4,2,6,3,1,5). To the right: At most three signals may be routed on the silicon surface without crossing each other. The dashed signals must be bridged.

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 pecifies 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.

InputOn 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.OutputFor 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

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题目很长（无聊的可以仔细看一下），大意就是从一端连线到另一端，要使得线没有交叉且连接数尽量大，所以是一道求最长递增子序列的题目（虽然集训队的学长人好已经在标题告诉我们了）。

这道题给了1s，只有nlogn的算法能过，那怎么办？

我们用一个数组来记录每个单调递增子序列，数组下标表示这个递增子序列的长度，该下标对应数组的值表示这个递增子序列的最大值。

比如说对于一个序列：5，3，4，2，7，初始化b[1]=a[1]=5，然后开始遍历原序列

1，a[2]=3<b[1]=5，就让b[1]=3；

2，a[3]=4>b[1]=3，那么b[2]=4;

3，a[4]=2<b[2]=4，那么从b数组里面找到比2大的最小的b[i]使得b[i]=2，所以b[1]=2;

4，a[5]=7>b[2]，那么b[3]=7；

于是最长递增子序列的长度就是3；

代码：

#include<iostream>
#include<cstdio>
using namespace std;

const int MAXN=4e4+1;
int a[MAXN],b[MAXN],m,n,len;
int bs(int left,int right,int a[],int val)//因为b数组是单调的，所以可以用二分搜索降低时间复杂度
{
    int r=right,l=left;
    while(l<r)
    {
        int mid=r+l>>1;
        if(val<a[mid])r=mid;
        else l=mid+1;
    }
    return l;
}
int main()
{
    scanf("%d",&m);
    for(int k=0;k<m;k++)
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++)scanf("%d",&a[i]);
        b[1]=a[1];
        len=1;
        for(int i=2;i<=n;i++)
        {
            if(a[i]>b[len])
            {
                len++;
                b[len]=a[i];
            }
            else{
                int index = bs(1,len,b,a[i]);//找到比a[i]大的最小的b数组元素的下标
                b[index]=a[i];
            }
        }
        printf("%d\n",len);
    }
}