洛谷 - P1020 导弹拦截 （LIS）

传送门：洛谷 - P1020

思路分析

第一问求最长不降序列
第二问求最长严格单升序列
其实把第一问反过来就是求最长上升子序列了

样例输入

389 207 155 300 299 170 158 65

样例输出

6
2

AC代码

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <bits/stdc++.h>
#define ls x<<1
#define rs x<<1|1
#define fi first
#define se second
#define ll long long
#define pb push_back
#define mp make_pair
#define fun function
#define sz(x) (x).size()
#define lowbit(x) (x)&(-x)
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))

namespace FastIO {
#define BUF_SIZE 100000
#define OUT_SIZE 100000
    bool IOerror=0;
    inline char nc() {
        static char buf[BUF_SIZE],*p1=buf+BUF_SIZE,*pend=buf+BUF_SIZE;
        if(p1==pend) {
            p1=buf;
            pend=buf+fread(buf,1,BUF_SIZE,stdin);
            if(pend==p1) {
                IOerror=1;
                return -1;
            }
        }
        return *p1++;
    }
    inline bool blank(char ch) {
        return ch==' '||ch=='\n'||ch=='\r'||ch=='\t';
    }
    template<class T> inline bool read(T &x) {
        bool sign=0;
        char ch=nc();
        x=0;
        for(; blank(ch); ch=nc());
        if(IOerror)return false;
        if(ch=='-')sign=1,ch=nc();
        for(; ch>='0'&&ch<='9'; ch=nc())x=x*10+ch-'0';
        if(sign)x=-x;
        return true;
    }
    template<class T,class... U>bool read(T& h,U&... t) {
        return read(h)&&read(t...);
    }
#undef OUT_SIZE
#undef BUF_SIZE
};
using namespace std;
using namespace FastIO;
using namespace __gnu_pbds;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> rbtree;

const int INF = 0x3f3f3f3f;
const int N = 1e5+10;

int a[N],b[N],cnt;

void add(int x,int val){
    for(int i=x;i<N;i+=lowbit(i)) b[i]=max(b[i],val);
}

int query(int x){
    int ans=0;
    for(int i=x;i;i-=lowbit(i)) ans=max(ans,b[i]);
    return ans;
}

int main() {

#ifdef xiaofan
    freopen("1.in","r",stdin);
    freopen("1.out","w",stdout);
#endif

    int x;
    while(read(x)) a[++cnt]=x;
    int ans1=0;
    for(int i=cnt;i>=1;i--){
        int sum=query(a[i])+1;
        ans1=max(ans1,sum);
        add(a[i],sum); 
    }
    mem(b,0);
    int ans2=0;
    for(int i=1;i<=cnt;i++){
        int sum=query(a[i]-1)+1;
        ans2=max(sum,ans2);
        add(a[i],sum);
    }
    printf("%d\n%d\n",ans1,ans2);




    return 0;
}