codeforce A - 1296C(1500)(map)_codeforce上考察map的题目

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

题目大意：
给你一个字符串，找出其中可以删掉的一段，要保证初始结尾位置不变。
题目分析：
用map套pair去储存每个点的坐标和是否出现过，然后遍历字符串，如果找到出现两次的坐标，则上次出现的坐标位置到该位置中间一段可以删去。而且还学到了make_pair(关于make_pair)
题解：

#include<bits/stdc++.h>
using namespace std;
map <pair<int,int>,int> mp; 
string s;
int n; 
int main(){
    int t;
    scanf("%d",&t);
    
    while(t--){
    	cin >> n;
    	cin >> s;
    	int _max = 1e9;
    	int x, y, l = 0, r = 0;
		mp.clear();
		mp[make_pair(0,0)] = -1;
		s=" "+s;
		
		for(int i=1; i<s.size(); i++){
			if(s[i]=='L') l--;
			else if(s[i]=='R') l++;
			else if(s[i]=='U') r++;
			else r--;
			
			int zz = mp[make_pair(l,r)]; //第一次出现是0  若是出现过 则为之前出现的位数 
			if(zz!=0){
				if(zz==-1) zz=0; //回到了（0，0）点 则设为0  
				if(i-zz<_max){  // 用该点在字符串中的位数 减去第一次出现的位数 如果存在的话 
					_max = i-zz;  //更新—最短的一段 
					x = zz+1; //删去的一段的左端 
					y = i;    //删去的一段的右段 
				}
			}
			mp[make_pair(l,r)] = i; //将该点的坐标赋给second 
			//将该点设为非零数 相当于已经走过了 
		}
		if(_max == 1e9) //没有被更新过 即没有重复出现过坐标 
			cout << -1 <<endl;
		else 
			cout << x << " " << y <<endl;
	}
    return 0;
}
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