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NC51218. Trip

描述

Alice and Bob want to go on holiday. Each of them has planned a route, which is a list of cities to be visited in a given order. A route may contain a city more than once. 
As they want to travel together, they have to agree on a common route. None wants to change the order of the cities on his or her route or add other cities. Therefore they have no choice but to remove some cities from the route. Of course the common route should be as long as possible. 
There are exactly 26 cities in the region. Therefore they are encoded on the lists as lower case letters from 'a' to 'z'.

输入描述

The input consists of two lines; the first line is Alice's list, the second line is Bob's list.
Each list consists of 1 to 80 lower case letters with no spaces inbetween.

输出描述

The output should contain all routes that meet the conditions described above, but no route should be listed more than once. Each route should be printed on a separate line. There is at least one such non-empty route, but never more than 1000 different ones. Output them in ascending order.

示例1

输入:

abcabcaa
acbacba

输出:

ababa
abaca
abcba
acaba
acaca
acbaa
acbca

原站题解

上次编辑到这里,代码来自缓存 点击恢复默认模板

C++(g++ 7.5.0) 解法, 执行用时: 4ms, 内存消耗: 688K, 提交时间: 2022-08-13 13:59:12

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef string str;
#define MAXN 211
ll f[MAXN][MAXN],la,lb;
char a[MAXN],b[MAXN];
ll loca[29][MAXN],locb[29][MAXN];
std::vector<str>res;
void printway(ll i,ll j,ll rest,str cur)
{
    if(!rest)
    {
        res.push_back(cur);
        return;
    }
    if(!i||!j)return;
    for(ll k=0;k<26;++k)
    {
        ll pi=loca[k][i],pj=locb[k][j];
        if(f[pi][pj]==rest)printway(pi-1,pj-1,rest-1,char('a'+k)+cur);
    }
}
int main()
{
    scanf("%s%s",a+1,b+1);
    la=strlen(a+1),lb=strlen(b+1);
    f[0][0]=0;
    for(ll i=1;i<=la;++i)
        for(ll j=1;j<=lb;++j)
        {
            ll cur=max(f[i-1][j],f[i][j-1]);
            if(a[i]==b[j])cur=f[i-1][j-1]+1;
            f[i][j]=cur;
        }
    for(ll i=1;i<=la;++i)
        for(ll j=0;j<26;++j)
            if(a[i]=='a'+j)loca[j][i]=i;
            else loca[j][i]=loca[j][i-1];
    for(ll i=1;i<=lb;++i)
        for(ll j=0;j<26;++j)
            if(b[i]=='a'+j)locb[j][i]=i;
            else locb[j][i]=locb[j][i-1];
    printway(la,lb,f[la][lb],"");
    std::sort(res.begin(),res.end());
    for(std::vector<str>::iterator it=res.begin();it!=res.end();++it)
        std::cout<<*it<<endl;
    return 0;
}

C++ 解法, 执行用时: 13ms, 内存消耗: 768K, 提交时间: 2022-01-30 14:25:44

#include<bits/stdc++.h>
using namespace std;
string s,t;
int f[1001][1001],n,m;
set<string>h;
void dfs(int k,int n,int m,string ss){
    if(!k){reverse(ss.begin(),ss.end());h.insert(ss);return;}
    for(int i=1;i<n;i++)for(int j=1;j<m;j++)if(s[i-1]==t[j-1]&&f[i][j]==k)ss.push_back(s[i-1]),dfs(k-1,i,j,ss),ss.pop_back();
}
int main(){
    cin>>s>>t;
    n=s.size(),m=t.size();
    for(int i=0;i<n;i++)for(int j=0;j<m;j++){f[i+1][j+1]=max(f[i+1][j],f[i][j+1]);if(s[i]==t[j]&&f[i][j]+1>=f[i+1][j+1])f[i+1][j+1]=f[i][j]+1;}
    dfs(f[n][m],n+1,m+1,"");
    for(auto it:h)cout<<it<<"\n";
    return 0;
}

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