#include<bits/stdc++.h>
using namespace std;
//input
#define rep(i,x,y) for(int i=(x);i<=(y);++i)
#define RI(n) scanf("%d",&(n))
#define RII(n,m) scanf("%d%d",&n,&m);
#define RIII(n,m,k) scanf("%d%d%d",&n,&m,&k)
#define RS(s) scanf("%s",s)
#define LL long long
#define REP(i,N)  for(int i=0;i<(N);i++)
#define CLR(A,v)  memset(A,v,sizeof A)
//
#define N 65
#define inf -0x3f3f3f3fint dp[100001];int dp2[100001];int a[100005];
int main()
{int x;int cnt=0;while(RI(x)==1)a[++cnt]=x;rep(i,1,cnt){dp[i]=1;dp2[i]=1;}int maxx=0;int minn=0;rep(i,2,cnt){   for(int j=1;j<i;j++){if(a[j]>=a[i])dp[i]=max(dp[i],dp[j]+1);maxx=max(dp[i],maxx);if(a[j]<a[i])dp2[i]=max(dp2[i],dp2[j]+1);minn=max(dp2[i],minn);}}cout<<maxx<<endl<<minn;return 0;
}

直接维护一个非升序列和一个非降序列，维护的过程就能用上面两个神奇的STL（当然非升序列的upper_bound()要重载cmp）。不多说，上代码#include<bits/stdc++.h>
using namespace std;
int a[100005],f[100005],l[100005];
struct cmp{bool operator()(int a,int b){return a>b;}};
int main()
{int n=1;while(cin>>a[n])n++;n--;int con=1,cont=1;l[1]=f[1]=a[1];for(int i=2;i<=n;i++){if(l[cont]>=a[i])l[++cont]=a[i];else l[upper_bound(l+1,l+cont+1,a[i],cmp())-l]=a[i];if(f[con]<a[i])f[++con]=a[i];else f[lower_bound(f+1,f+con+1,a[i])-f]=a[i];}cout<<cont<<" "<<con;return 0;
}

#include<bits/stdc++.h>
using namespace std;
//input
#define rep(i,x,y) for(int i=(x);i<=(y);++i)
#define RI(n) scanf("%d",&(n))
#define RII(n,m) scanf("%d%d",&n,&m);
#define RIII(n,m,k) scanf("%d%d%d",&n,&m,&k)
#define RS(s) scanf("%s",s)
#define LL long long
#define REP(i,N)  for(int i=0;i<(N);i++)
#define CLR(A,v)  memset(A,v,sizeof A)
//
#define N 100005
#define inf -0x3f3f3f3fint c[N];
int lowbit(int x)
{return x&(-x);
}void update(int x,int v)
{for(int i=x;i<=N;i+=lowbit(i))c[i]=max(c[i],v);
}
int sum(int x)
{int ans=0;for(int i=x;i>0;i-=lowbit(i))ans=max(ans,c[i]);return ans;
}int a[N];
int main()
{int x;int cnt=0;while(RI(x)==1)a[++cnt]=x;int ans1=0,ans2=0;for(int i=cnt;i>=1;i--){int p=sum(a[i])+1;update( a[i],p);ans1=max(ans1,p);}cout<<ans1<<endl;CLR(c,0);rep(i,1,cnt){int p=sum(a[i]-1 )+1;update(a[i],p);ans2=max(ans2,p);}cout<<ans2;return 0;
}