题解 [CSP-S 2021] 回文

传送门

• 多测题里不要写 exit(0);

四个指针模拟，贪心选数
尽量优先选左边
枚举第一次选什么
数两两不同，所以没有后效性
为了形成回文，选第 \(i\) 个数的时候第 \(2n-i+1\) 个数已经确定了，看看能不能选就行了
不能就-1

Code:

#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define N 1000010
#define ll long long
#define pb push_back
//#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];

namespace force{
	vector< vector<char> > ans;
	vector<char> t;
	int tem[N], top;
	bool check(int s) {
		top=0;
		int pos1=1, pos2=n*2;
		for (int i=0; i<n*2; ++i) {
			if (s&(1<<i)) tem[++top]=a[pos2--];
			else tem[++top]=a[pos1++];
		}
		for (int i=1; i<=n; ++i) if (tem[i]!=tem[n*2-i+1]) return 0;
		return 1;
	}
	void solve() {
		ans.clear();
		int lim=1<<(n*2);
		for (int s=0; s<lim; ++s) {
			if (check(s)) {
				t.clear();
				for (int i=0; i<n*2; ++i) {
					if (s&(1<<i)) t.push_back('R');
					else t.push_back('L');
				}
				ans.push_back(t);
			}
		}
		sort(ans.begin(), ans.end());
		if (!ans.size()) {printf("-1\n"); return ;}
		for (int i=0; i<(int)ans[0].size(); ++i) printf("%c", ans[0][i]);
		printf("\n");
	}
}

#if 0
namespace task1{
	vector<char> t1, t2;
	int 
	void calc(int l1, int r1, int l2, int r2, vector<int>& v) {
		
	}
	void solve() {
		t1.clear(); t2.clear();
		for (int i=2; i<=n*2; ++i) if (a[i]==a[1]) {
			calc(2, i-1, i+1, n*2, t1);
			break;
		}
		for (int i=1; i<n*2; ++i) if (a[i]==a[n]) {
			calc(1, i-1, i+1, n*2-1, t2);
			break;
		}
		if (t1>t2) swap(t1, t2);
		for (int i=0; i<int(t1.size()); ++i) printf("%c", t1[i]);
		printf("\n");
	}
}
#endif

namespace task2{
	bool vis[N];
	void solve() {
		for (int i=1; i<=n; ++i) {
			if (vis[a[i]]) {
				for (int j=1; j<=i; ++j) vis[a[j]]=0;
				puts("-1");
				return ;
			}
			else vis[a[i]]=1;
		}
		for (int i=1; i<=n; ++i) vis[a[i]]=0;
		for (int i=1; i<=n; ++i) printf("L");
		for (int i=1; i<n; ++i) printf("R");
		puts("L");
	}
}

namespace task{
	vector<char> t1, t2, t;
	bool calc(int pos1, int pos2, int pos3, int pos4, vector<char>& v) {
		t.clear();
		while (1) {
			if (pos1<pos2 && a[pos1]==a[pos2]) v.pb('L'), t.pb('L'), ++pos1, --pos2;
			else if (pos1<=pos2 && pos3<=pos4 && a[pos1]==a[pos3]) v.pb('L'), t.pb('R'), ++pos1, ++pos3;
			else if (pos3<pos4 && a[pos3]==a[pos4]) v.pb('R'), t.pb('R'), ++pos3, --pos4;
			else if (pos1<=pos2 && pos3<=pos4 && a[pos2]==a[pos4]) v.pb('R'), t.pb('L'), --pos2, --pos4;
			else if (pos1>pos2 && pos3>pos4) break;
			else return 0;
		}
		for (int i=t.size()-1; ~i; --i) v.pb(t[i]);
		return 1;
	}
	void solve() {
		t1.clear(); t2.clear();
		int pos;
		for (int i=2; i<=n*2; ++i) if (a[i]==a[1]) {pos=i; break;}
		if (calc(2, pos-1, pos+1, n*2, t1)) {
			printf("L"); for (auto it:t1) putchar(it); puts("L");
			return ;
		}
		for (int i=1; i<n*2; ++i) if (a[i]==a[n*2]) {pos=i; break;}
		if (calc(1, pos-1, pos+1, n*2-1, t2)) {
			printf("R"); for (auto it:t2) putchar(it); puts("L");
			return ;
		}
		puts("-1");
	}
}

signed main()
{
	//freopen("palin.in", "r", stdin);
	//freopen("palin.out", "w", stdout);

	int T=read();
	while (T--) {
		n=read();
		for (int i=1; i<=n*2; ++i) a[i]=read();
		// if (n<=10) force::solve();
		// else task2::solve();
		task::solve();
	}

	return 0;
}