【openjudge 1759】最长上升子序列（dp）

【题解】【O(n²)&O(nlogn)】

朴素最长上升子序列O(n²)：

#include
#include
#include
using namespace std;
int f[10010],n,a[10010];
int main()
{
	int i,j;
	scanf("%d",&n);
	for (i=1;i<=n;i++)
	 scanf("%d",&a[i]);
	for (i=1;i<=n;i++)
	  f[i]=1;
	for (i=2;i<=n;i++)
	 for (j=1;jf[i])
	     f[i]=f[j]+1;
	for (i=2;i<=n;i++)
	  f[i]=max(f[i-1],f[i]);
	printf("%d",f[n]);
	return 0;
} 

用树状数组维护最长上升子序列O(nlogn)：

#include
#include
#include
using namespace std;
int tree[100010],a[10010],num[10010],n,ans;
inline int lowbit(int x)
{
	return (x&(-x));
}
inline void add(int x,int val)
{
	for(int i=x;i<=n;i+=lowbit(i))
	 if(val>tree[i]) tree[i]=val;
}
inline int ask(int x)
{
	int sum=0;
	for(int i=x;i>0;i-=lowbit(i))
	 if(tree[i]>sum) sum=tree[i];
	return sum;
}
int main()
{
	int i,j;
	scanf("%d",&n);
	for(i=1;i<=n;++i)scanf("%d",&a[i]),num[i]=a[i];
	sort(num+1,num+n+1);
	int m=unique(num+1,num+n+1)-num-1;
	for(i=1;i<=n;++i)
	 {
	 	a[i]=lower_bound(num+1,num+m+1,a[i])-num;
	 	int tmp=ask(a[i])+1;
	 	if(tmp>ans) ans=tmp;
	 	add(a[i]+1,tmp);
	 }
	printf("%d\n",ans);
	return 0;
}