Yet Another Walking Robot（奇妙的记录方式）

There is a robot on a coordinate plane. Initially, the robot is located at the point (0,0). Its path is described as a string s of length n consisting of characters ‘L’, ‘R’, ‘U’, ‘D’.

Each of these characters corresponds to some move:

‘L’ (left): means that the robot moves from the point (x,y) to the point (x−1,y);
‘R’ (right): means that the robot moves from the point (x,y) to the point (x+1,y);
‘U’ (up): means that the robot moves from the point (x,y) to the point (x,y+1);
‘D’ (down): means that the robot moves from the point (x,y) to the point (x,y−1).
The company that created this robot asked you to optimize the path of the robot somehow. To do this, you can remove any non-empty substring of the path. But this company doesn’t want their customers to notice the change in the robot behavior. It means that if before the optimization the robot ended its path at the point (xe,ye), then after optimization (i.e. removing some single substring from s) the robot also ends its path at the point (xe,ye).

This optimization is a low-budget project so you need to remove the shortest possible non-empty substring to optimize the robot’s path such that the endpoint of his path doesn’t change. It is possible that you can’t optimize the path. Also, it is possible that after the optimization the target path is an empty string (i.e. deleted substring is the whole string s).

Recall that the substring of s is such string that can be obtained from s by removing some amount of characters (possibly, zero) from the prefix and some amount of characters (possibly, zero) from the suffix. For example, the substrings of “LURLLR” are “LU”, “LR”, “LURLLR”, “URL”, but not “RR” and “UL”.

You have to answer t independent test cases.

Input
The first line of the input contains one integer t (1≤t≤1000) — the number of test cases.

The next 2t lines describe test cases. Each test case is given on two lines. The first line of the test case contains one integer n (1≤n≤2⋅105) — the length of the robot’s path. The second line of the test case contains one string s consisting of n characters ‘L’, ‘R’, ‘U’, ‘D’ — the robot’s path.

It is guaranteed that the sum of n over all test cases does not exceed 2⋅105 (∑n≤2⋅105).

Output
For each test case, print the answer on it. If you cannot remove such non-empty substring that the endpoint of the robot’s path doesn’t change, print -1. Otherwise, print two integers l and r such that 1≤l≤r≤n — endpoints of the substring you remove. The value r−l+1 should be minimum possible. If there are several answers, print any of them.

Example
Input
4
4
LRUD
4
LURD
5
RRUDU
5
LLDDR
Output
1 2
1 4
3 4
-1
map学的还是不行啊，不说了，给上依靠大佬的代码。

#include<iostream>
#include<bits/stdc++.h>
#include<algorithm>
#include<math.h>
#include<string.h>
#include<string>
#include<map> 
using namespace std;
const int N = 1e6+10;
map<pair<int,int>,int> mp;
//前面是点的位置，最近一次走过此点的指令位置
int a[N];
int n , m;
int main()
{
	int n , t;
	string s;
	cin >> t;
	while(t--)
	{
		cin >> n >> s;
		int x=0 , y=0 , l=0 , r=n;
		mp.clear();
		mp[{0,0}]=1;
		//起始位置，也算走过一次，所以赋值为1
		for(int i = 0 ; i < s.size() ; i++)
		{
			if(s[i]=='L')
			  x--;
			else if(s[i]=='R')
			  x++;
			else if(s[i]=='U')
			  y++;
			else
			  y--;
			if(mp[{x,y}]!=0&&r-l>i-mp[{x,y}]+1)
			{
				//如果此点之前走过并且指令的长度比之前储存的长度要短
				l=mp[{x,y}];
				//左边界就是之前的位置
				r=i+1;
				//右边界是i+1,因为l是从1开始，比i大一
			}
			mp[{x,y}]=i+2;
			//i是从0开始的，第一个指令之后应该是第二个，所以可见是i+2
		}
		if(l==0)
		{
			cout << -1 << endl;
		}
		else
		{
			cout << l << ' ' << r << endl;
		}
	}
	return 0;
}