Topic link: "Blue Bridge Cup" Practice System
Analysis: This is a longest ascending subsequence problem, except that each element is not an integer but a string, but the essence is the same. For the longest ascending subsequence problem, I will give two methods below.
I will not go into details about the problem of solving the longest ascending subsequence. I have already covered it in detail in the previous blog. For details, please refer to: Detailed explanation of the longest subsequence problem
The first method is ordinary one-dimensional dp, f[i] represents the maximum length of the longest ascending subsequence ending with the i-th string. I will not talk about the dynamic transition equation. What I want to say is don’t forget Open a pre array to record what is the previous position transferred to the current position. This is updated during the process of updating f, but this can only pass 70% of the samples. After all, the complexity is o(n^2). , the code is attached below:
#include<cstdio>
#include<iostream>
#include<cstring>
#include<vector>
#include<algorithm>
#include<map>
#include<cmath>
#include<queue>
using namespace std;
const int N=1000003;
string s[N];
int f[N],pre[N],st[N];//f[i]表示以第i个名字结尾的最长上升子序列
int t1[N],t2[N];
bool cmp(int i,int j)//判断以第i个名字结尾的上升子序列和以第j个名字结尾的上升子序列谁字典序更小
{
if(f[i]<f[j]) return true;
if(f[i]>f[j]) return false;
int tt1=0,tt2=0;
while(i)
{
t1[++tt1]=i;
i=pre[i];
}
while(j)
{
t2[++tt2]=j;
j=pre[j];
}
for(int i=tt1;i>=1;i--)
if(s[t1[i]]<s[t2[i]]) return false;
else if(s[t1[i]]>s[t2[i]]) return true;
}
int main()
{
int tt=0;
string p;
cin>>p;
for(int i=0;i<p.size();i++)
{
if(p[i]<='Z'&&p[i]>='A')
s[++tt]+=p[i];
else
s[tt]+=p[i];
}
int ans=0;
for(int i=1;i<=tt;i++)
{
f[i]=1;
for(int j=1;j<i;j++)
{
if(s[i]>s[j])
{
if(f[i]<f[j]+1)
{
f[i]=f[j]+1;
pre[i]=j;//记录路径
}
else if(f[i]==f[j]+1)
{
if(s[pre[i]]>s[j])
{
pre[i]=j;//记录路径
}
}
}
if(cmp(ans,i))
ans=i;
}
}
tt=0;
while(ans)
{
st[++tt]=ans;
ans=pre[ans];
}
for(int i=tt;i;i--)
cout<<s[st[i]];
return 0;
}The following is the optimized code of the monotonic stack. The complexity is o(nlogn). I have given the proof in the blog attached above. It also includes how to output the final monotonic ascending subsequence. It is written in detail. You can read it. For a moment, the code is given directly below:
#include<cstdio>
#include<iostream>
#include<cstring>
#include<vector>
#include<algorithm>
#include<map>
#include<cmath>
#include<queue>
using namespace std;
const int N=1000003;
string s[N];
int st[N];
vector<string> ans;
int id[N];//id[i]记录第i个数在最长上升子序列中的位置
int main()
{
int tt=0;
string p;
cin>>p;
for(int i=0;i<p.size();i++)
{
if(p[i]<='Z'&&p[i]>='A')
s[++tt]+=p[i];
else
s[tt]+=p[i];
}
ans.push_back(s[1]);
int id_len=0;id[++id_len]=1;
for(int i=2;i<=tt;i++)
{
if(s[i]>ans[ans.size()-1])
{
ans.push_back(s[i]);
id[++id_len]=ans.size();
continue;
}
int t=lower_bound(ans.begin(),ans.end(),s[i])-ans.begin();
ans[t]=s[i];
id[++id_len]=t+1;
}
tt=0;
for(int i=ans.size(),j=id_len;i>=1;j--)
if(id[j]==i)
st[++tt]=j,i--;
for(int i=tt;i>=1;i--)
cout<<s[st[i]];
return 0;
}