ARC-195-A - Twice Subsequence · znzryb

## 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

```cpp
// 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](https://atcoder.jp/contests/arc195/submissions/64142795)