HDU 1950 Bridging signals

http://acm.hdu.edu.cn/showproblem.php?pid=1950

和上一题基本一样，二分dp求LIS

ps：排名居然刷到第一了，原来从来没有遇到呢，好开心~

#include <iostream>
using namespace std ;
int dp[10001],p[40001];
int LIS(int n)
{
    int l,r,m,i,tail = 0;
    for ( dp[ ++ tail ] = p[ 1 ],i = 2 ; i <= n ; ++ i ) 
    {
        if ( dp[ tail ] <= p[ i ] ) 
        {
            dp[ ++ tail ] = p[ i ];
            continue;
        }
        for ( m=((r=tail)+(l=1)>>1) ; l < r ; m=(l+r)>>1 )
            if ( dp[ m ] <= p[ i ] ) l = m+1;
            else r = m;
        dp[ m ] = p[ i ];
    }
    return tail;
}
inline bool scan_d(int &num) 
{
        char in;bool IsN=false;
        in=getchar();
        if(in==EOF) return false;
        while(in!='-'&&(in<'0'||in>'9')) in=getchar();
        if(in=='-'){ IsN=true;num=0;}
        else num=in-'0';
        while(in=getchar(),in>='0'&&in<='9'){
                num*=10,num+=in-'0';
        }
        if(IsN) num=-num;
        return true;
}
int main()
{
    int t,n,nCase=1;
    scan_d(t);
    while(t--)
    {
        scan_d(n);
        for(int i=1;i<=n;i++)
        {
            int x;
            scan_d(x);
            p[i]=x;
        }
        int ans=LIS(n);
        printf("%d\n",ans);
    }
    return 0;
}