B – Bridging signals （LIS）

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.

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

错误代码（一组数据行，怎么实现多组数据输入？）

Select Code

#include <stdio.h>
#include <cstring>
#include <algorithm>
#define INF 0x3f3f3f
using namespace std;
int dp[30020],a[30020];
int main()
{
int n;
scanf("%d",&n);
while(n--)
{
int m;
scanf("%d",&m);
for(int i=0; i<m ; i++)
{
scanf("%d",&a[i]);
dp[i]=INF;
}
for(int i=0; i<m; i++)
*lower_bound(dp,dp+n,a[i])=a[i];//优化
printf("%d\n",lower_bound(dp,dp+m,INF)-dp);
}
return 0;
}

正确代码

Select Code

#include<stdio.h>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std; int a[41000];
int main()
{
int N,n,i,j,t;
scanf("%d",&N);
while(N--)
{
scanf("%d",&n);
scanf("%d",&t);
a[0]=t;
int top=1;
for(i=1; i<n; i++)
{
scanf("%d",&t);
if(t>=a[top-1])
a[top++]=t;
else
{
int z=0,q=top-1;
while(z<=q)
{
int mid=(z+q)/2;
if(a[mid]<t)
z=mid+1;
else
q=mid-1;
}
a[z]=t;
}
}
printf("%d\n",top);
}
return 0;
}