题解 [CF1672D] Cyclic Rotation

传送门

一眼感觉比较神，但性质还是很好找的
发现可以倒着扫 b 数组，需要某个数的话倒着从 a 中取
是需要的颜色就取，否则需要换到后面去
如果后面不需要它则无解

点击查看代码

#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define N 200010
#define ll long long
//#define int long long

char buf[1<<21], *p1=buf, *p2=buf;
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf, 1, 1<<21, stdin)), p1==p2?EOF:*p1++)
inline int read() {
	int ans=0, f=1; char c=getchar();
	while (!isdigit(c)) {if (c=='-') f=-f; c=getchar();}
	while (isdigit(c)) {ans=(ans<<3)+(ans<<1)+(c^48); c=getchar();}
	return ans*f;
}

int n;
int a[N], b[N];
map<int, int> mp;

signed main()
{
	int T=read();
	while (T--) {
		n=read();
		mp.clear();
		for (int i=1; i<=n; ++i) a[i]=read();
		for (int i=1; i<=n; ++i) b[i]=read();
		int p1, p2;
		for (p1=n,p2=n; p2; --p2) {
			int k=1;
			while (p2 && b[p2-1]==b[p2]) --p2, ++k;
			while (p1>0 && a[p1]!=b[p2]) {
				if (mp[a[p1]]<0) ++mp[a[p1--]];
				else {puts("NO"); goto jump;}
			}
			if (p1<=0) {puts("NO"); goto jump;}
			else mp[a[p1--]]-=k-1;
		}
		while (p1>0) ++mp[a[p1--]];
		for (auto it:mp) if (it.second) {puts("NO"); goto jump;}
		puts("YES");
		jump: ;
	}
	
	return 0;
}