HDU 1950 Bridging signals(LIS nlogn)

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 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.

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

题目大意：有平行的两列信号点，信号点之间有一些连线，有的交叉，现在不允许交叉，求最多能留下几根线 。

题解：思路和题目 Constructing Roads In JGShining’s Kingdom差不多。此题是求出最长上升子序列里的元素个数，以某一边为序，求一个最长上升子序列的长度即可，普通方法可能超时，可以用二分法，这里用了函数lower_bound( ) ，和二分法意图一样，个人感觉更好用，时间缩短为nlogn，具体实现看代码：

函数知识的一点用法补充：
ForwardIter lower_bound(ForwardIter first, ForwardIter last,const _Tp& val)算法返回一个非递减序列[first, last)中的第一个大于等于值val的位置。

ForwardIter upper_bound(ForwardIter first, ForwardIter last, const _Tp& val)算法返回一个非递减序列[first, last)中第一个大于val的位置。

lower_bound和upper_bound如下图所示：

实际运用中的写法是 k = lower_bound(a,a+n,val)-a（a为数组a[]，里面有n个元素）为什么在末尾减a,可以想一下

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;#define INF 0x3f3f3f3fint v[40010],li[40010];//v[]储存最长上升子序列int main(){    int n,p,i,k,ans;    scanf("%d",&n);    while(n--)    {        scanf("%d",&p);        for(i=1;i<=p;i++)        {            scanf("%d",&li[i]);            v[i]=INF;//初始化，v[]里的元素都为无穷大，以便后边进行比较        }        ans=0;        for(i=1;i<=p;i++)        {            k=lower_bound(v+1,v+1+p,li[i])-v;//在v[]中找到第一个        不小于li[i]的数，返回值k为此数的位置下标；            v[k]=min(v[k],li[i]);//若li[i]比v[k]小，则进行替换更新；            ans=max(ans,k);//每次与k比较，记录序列当前的最大长度        }        printf("%d\n",ans);    }       return 0;}