Bridging signals （POJ 1631 HDU 1950）

'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

题解：就是求最长上升子序列，正常套用模板会T，可以改用二分或者使用lower_bound。

代码如下：

#include <iostream>
#include <cstdio>
#include <stdlib.h>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <string.h>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <ctime>
#define maxn 1007
#define N 100005
#define INF 0x3f3f3f3f
#define PI acos(-1)
#define lowbit(x) (x&(-x))
#define eps 0.000000001
#define read(x) scanf("%d",&x)
#define put(x) printf("%d\n",x)
#define Debug(x) cout<<x<<" "<<endl
using namespace std;
typedef long long ll;

int a[40007],dp[40007],n;

int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        memset(a,0,sizeof(a));
        memset(dp,0,sizeof(dp));
        cin>>n;
        for(int i=1; i<=n; i++)
        {
            cin>>a[i];
        }
        int top=0;
        for(int i=1;i<=n;i++)
		{
			if(a[i]>dp[top])
				dp[++top]=a[i];
			else
			{
				int pos=lower_bound(dp+1,dp+top+1,a[i])-dp;
				dp[pos]=a[i];
			}
		}
        cout<<top<<endl;
    }
    return 0;
}

二分（某大佬代码）：

#include <cstdio>
#include <cstring>
#include <algorithm>
#define MAXN 40000+10
#define INF 1000000;
using namespace std;
//首先我们用dp[i]表示以第i个元素结尾的LIS的长度
int N;
int a[MAXN];
int s[MAXN];//s[k]存储满足dp[i] = k时所有a[i]里面的最小值
int binary_search(int *s, int now, int length)
{
    int left = 0, right = length, mid;
    while(right >= left)
    {
        mid = (left + right) >> 1;
        if(s[mid] == now)//相等返回数组对应下标
            return mid;
        else if(s[mid] > now)//比它大
            right = mid - 1;
        else
            left = mid + 1;
    }
    return left;
}
int main()
{
    int t;
    int len;//存储长度
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d", &N);
        for(int i = 1; i <= N; i++)
            scanf("%d", &a[i]);
        s[0] = -1;
        len = 1;
        for(int i = 1; i <= N; i++)
        {
            s[len] = INF;//最后一位设置无穷大
            //在数组s里面 寻找第一个比a[i]大的元素的位置
            int j = binary_search(s, a[i], len);
            if(j == len)//比s数列里面所有元素都要大
                len++;
            s[j] = a[i];//替换
        }
        //到最后数组s里面元素的个数就是LIS的长度
        printf("%d\n", len - 1);
    }
    return 0;
}