HDU4604.Deque——nlogn求最长上升子序列的长度

http://acm.hdu.edu.cn/showproblem.php?pid=4604

把一个序列中的元素放到队列里面，有3种操作，对于第i个元素
1.放到队头2.放到队尾3.舍弃
求最长上升序列的长度

法1:求出每个以a[i]为起点的最长不升子序列的长度，和最长不降子序列的长度，两个相加再减去a[i]重复的次数

法2:把当前序列复制两个，一个逆序，求总共的最长上升子序列的长度，然后奇偶避免重复

//法1
#include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
#include <algorithm>
#define foreach(v) (v).begin(),(v).end()
#define pk push_back
const int maxn = 100010;
using namespace std;
int num[maxn];
int down[maxn],up[maxn],down_same[maxn],up_same[maxn];
int n;
vector<int> vi;
void solve(int dp[],int same[])
{
    vi.clear();
    for(int i=n;i>=1;--i){
        int a=lower_bound(foreach(vi),num[i])-vi.begin();
        int b=upper_bound(foreach(vi),num[i])-vi.begin();
        if(b==vi.size()||vi.empty()){
            vi.pk(num[i]);
            dp[i]=vi.size();
        }else{
            vi[b]=num[i];
            dp[i]=b+1;
        }
        same[i]=b-a+1;
    }
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("in.txt","r",stdin);
freopen("out.txt","w",stdout);
#endif // ONLINE_JUDGE
    int T;
    scanf("%d",&T);
    while(T--){
        scanf("%d",&n);
        for(int i=1;i<=n;++i){
            scanf("%d",&num[i]);
        }
        solve(down,down_same);
        for(int i=1;i<=n;++i){
            num[i]=-num[i];
        }
        solve(up,up_same);
        int ans=0;
        for(int i=1;i<=n;++i){
            ans=max(ans,up[i]+down[i]-min(up_same[i],down_same[i]));
        }
        printf("%d\n",ans);
    }
    return 0;
}

//法2
#include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
#include <algorithm>
#define foreach(v) (v).begin(),(v).end()
#define pk push_back
const int maxn = 100010;
using namespace std;
int a[maxn];
int n;
vector<int> vi;
void solve(int x)
{
    int b=upper_bound(foreach(vi),x)-vi.begin();
    if(b==vi.size()||vi.empty())
        vi.pk(x);
    else
        vi[b]=x;
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("in.txt","r",stdin);
freopen("out.txt","w",stdout);
#endif // ONLINE_JUDGE
    int T;
    scanf("%d",&T);
    while(T--){
        scanf("%d",&n);
        vi.clear();
        for(int i=1;i<=n;++i){
            scanf("%d",&a[i]);
        }
        for(int i=n;i>=1;--i){
            solve(2*a[i]+1);
        }
        for(int i=1;i<=n;++i){
            solve(2*a[i]);
        }
        printf("%d\n",vi.size());
    }
    return 0;
 }