南阳214----单调递增子序列(二)

/*
用一个数组记录递增子序列，保持s处为最长子序列的最后一个值
当输入x小于d[s]时，向前找x的位置覆盖即可
复杂度与经典算法同为n*n
加入二分查找，优化后为n*logn
*/
#include<cstdio>
#define inf 1<<30
int d[100005],s;

void solven(int x)
{
    int t,left,right,mid;
    if(s == 0 || d[s] < x)
        d[++s] = x;
    else
    {
        left = 1;
        right = s;
        while(left <= right)
        {
            mid = (left + right)/2;
            if(d[mid]>=x && d[mid-1]<x)
            {
                d[mid] = x;
                return ;
            }
            if(d[mid] > x)
                right = mid - 1;
            else
                left = mid + 1;
        }
    }
}

int main()
{
    int n,x,i,t;
    while(~scanf("%d",&n))
    {
        s = 0;
        d[0] = -inf;
        while(n--)
        {
            scanf("%d",&x);
            solven(x);
        }
        printf("%d\n",s);
    }
    return 0;
}