poj 1631 Bridging signals【最长上升子序列】

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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 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.

Input

Output

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

解题思路：用暴力解决超时，于是在查找插入的时候采用2分法，轻松解决问题。

#include<stdio.h>
int a[40000],t,n,num,res,i,w;
int Q(int left,int right,int path){
    int mid;
    while(left<=right){
        mid=(left+right)/2;
        if(a[mid]<path) left=mid+1;
        else  right=mid-1;
   }return right;
}
int main(){
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        scanf("%d",&num);
        a[0]=num;
        res=1;
        for(i=1;i<n;i++){
            scanf("%d",&num);
            if(num>a[res-1]) a[res++]=num;
            else if(num<a[res-1]){
                w=Q(0,res-1,num);
                a[w+1]=num;
            }
        }
        printf("%d\n",res);
    }return 0;
}

#include<stdio.h>
#include<algorithm>
using namespace std;
int a[40000],t,n,num,res,i,w;
int main(){
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        scanf("%d",&num);
        a[0]=num;
        res=1;
        for(i=1;i<n;i++){
            scanf("%d",&num);
            if(num>a[res-1]) a[res++]=num;
            else if(num<a[res-1]){
                w=lower_bound(a,a+res,num)-a;
                a[w]=num;
            }
        }
        printf("%d\n",res);
    }return 0;
}

STL最简版

#include<stdio.h>
#include<algorithm>
using namespace std;
int a[40000],t,n,num,i,k,size;
int main(){
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        for(i=0,size=0;i<n;i++){
            scanf("%d",&num);
            k=lower_bound(a,a+size,num)-a;
            a[k]=num;
            size=max(size,k+1);
        }
        printf("%d\n",size);
    }return 0;
}