HDU 1950~Bridging signals~二分法求解

Bridging signals

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 981    Accepted Submission(s): 641

Problem 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

4642631510234567891018876543219589231746

 

Sample Output

3914

 

学长讲的时候一直没搞懂为什么会用二分法，后来才慢慢想通的～这道题呢可以找规律～可知要求输出的是最长上升子序列的个数～当你找最长上升子序列的时候可能会遇到很多个上升子序列，但你并不知道哪个是最长的~然后就需要用到二分法了～～～

定义b[k]:长度为k的上升子序列的最末元素，若有多个长度为k的上升子序列，则记录最小的那个最末元素。

注意d中元素是单调递增的，下面要用到这个性质。

首先len = 1,b[1] = a[1],然后对a[i]:若a[i]>b[len]，那么len++，b[len] = a[i];

否则，我们要从d[1]到d[len-1]中找到一个j，满足b[j-1]<a[i]<b[j],则根据题义，我们需要更新长度为j的上升子序列的最末元素（使之为最小的）即 b[j] = a[i];

最终答案就是len

利用b的单调性，在查找j的时候可以二分查找，从而时间复杂度为nlogn。

#include<stdio.h>#define N 40010int a[N],b[N];int main(){int n,m,i,len,r,l,mid;scanf("%d",&n);while(n--){len=1;scanf("%d",&m);for(i=1;i<=m;i++)scanf("%d",&a[i]);b[1]=a[1];for(i=2;i<=m;i++){if(a[i]>b[len])b[++len]=a[i];else{l=1;r=len;while(l<=r){mid=(r+l)/2;if(b[mid]<a[i])l=mid+1;elser=mid-1;}b[l]=a[i];}}printf("%d\n",len);}return 0;}