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ARC-195-A - Twice Subsequence

思路讲解

这种匹配子序列的可以用map+二分做,也可以用naive暴力

如果只是要看有没有两个的话,其实可以找出一个标号往大里走的(通过反向遍历得到),另一个标号往小里走的,对一对看看是不是一样的。

AC代码

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// Problem: A - Twice Subsequence
// Contest: AtCoder - AtCoder Regular Contest 195 (Div. 2)
// URL: https://atcoder.jp/contests/arc195/tasks/arc195_a
// Memory Limit: 1024 MB
// Time Limit: 2000 ms
// by znzryb
//
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (int i = (a); i >= (b); --i)
#define all(x) x.begin(),x.end()
#define CLR(i,a) memset(i,a,sizeof(i))
#define fi first
#define se second
#define pb push_back
#define SZ(a) ((int) a.size())

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll,ll> pll;
typedef array<ll,3> arr;
typedef double DB;
typedef pair<DB,DB> pdd;
typedef pair<ll,bool> plb;
constexpr ll MAXN=static_cast<ll>(1e6)+10,INF=static_cast<ll>(5e18)+3;

ll N,M,T,A[MAXN],B[MAXN];

inline void solve(){
cin>>N>>M;
FOR(i,1,N){
cin>>A[i];
}
FOR(i,1,M){
cin>>B[i];
}
vector<ll> seq;
ll idx=1;
FOR(i,1,N){
if(idx>M){
break;
}
if(A[i]==B[idx]){
seq.pb(i);
++idx;
}
}
if(idx<=M){
cout<<"No\n";
return;
}
vector<ll> reSeq;
idx=M;
ROF(i,N,1){
if(idx<=0){
break;
}
if(A[i]==B[idx]){
reSeq.pb(i);
--idx;
}
}
reverse(all(reSeq));
FOR(i,0,M-1){
if(seq[i]!=reSeq[i]){
cout<<"Yes\n";
return;
}
}
cout<<"No\n";
}

int main()
{
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
// cin>>T;
// while(T--){
// solve();
// }
solve();
return 0;
}
/*
AC
https://atcoder.jp/contests/arc195/submissions/64144783
*/

心路历程(WA,TLE,MLE……)

WA,没查出问题出在哪了

https://atcoder.jp/contests/arc195/submissions/64142795