2013
11-10

# Bridging signals

‘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.

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.

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.

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

3
9
1
4

//* @author:
import java.util.*;
public class Main {
static int a[]=new int[40000];
static int b[]=new int[40000];

static int binary(int i,int k){
if(a[i]< b[1])return 1;
int h=1,j=k;
while(h!=j-1){
k=(h+j)/2;
if(b[k]<=a[i])h=k;
else
j=k;
}
return j;
}
public static void main(String[] args) {
Scanner in=new Scanner(System.in);

int n,p,k,i;
n=in.nextInt();
while((n--)!=0){
p=in.nextInt();
for(i=0;i< p;i++)
a[i]=in.nextInt();
b[1]=a[0];
k=1;
for(i=1;i< p;i++){
if(a[i]>=b[k])b[++k]=a[i];
else b[binary(i,k)]=a[i];
}
System.out.println(k);
}
}
}

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2. 我还有个问题想请教一下，就是感觉对于新手来说，递归理解起来有些困难，不知有没有什么好的方法或者什么好的建议？