There is a robot on a coordinate plane. Initially, the robot is located at the point (0,0)(0,0) . Its path is described as a string ss of length nn 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)(x,y) to the point (x−1,y)(x−1,y) ;
- 'R' (right): means that the robot moves from the point (x,y)(x,y) to the point (x+1,y)(x+1,y) ;
- 'U' (up): means that the robot moves from the point (x,y)(x,y) to the point (x,y+1)(x,y+1) ;
- 'D' (down): means that the robot moves from the point (x,y)(x,y) to the point (x,y−1)(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)(xe,ye) , then after optimization (i.e. removing some single substring from ss ) the robot also ends its path at the point (xe,ye)(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 ss ).
Recall that the substring of ss is such string that can be obtained from ss 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 tt independent test cases.
The first line of the input contains one integer tt (1≤t≤10001≤t≤1000 ) — the number of test cases.
The next 2t2t lines describe test cases. Each test case is given on two lines. The first line of the test case contains one integer nn (1≤n≤2⋅1051≤n≤2⋅105 ) — the length of the robot's path. The second line of the test case contains one string ss consisting of nn characters 'L', 'R', 'U', 'D' — the robot's path.
It is guaranteed that the sum of nn over all test cases does not exceed 2⋅1052⋅105 (∑n≤2⋅105∑n≤2⋅105 ).
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 ll and rr such that 1≤l≤r≤n1≤l≤r≤n — endpoints of the substring you remove. The value r−l+1r−l+1 should be minimum possible. If there are several answers, print any of them.
4 4 LRUD 4 LURD 5 RRUDU 5 LLDDR
1 2 1 4 3 4 -1
數據結構題，考stl的map怎么用。用map<pair<int,int>,int>存儲走過的地方，是坐標點到字符串下標的映射。只要發現下一個走到的點已經出現在字典里了，說明這一段走的路就是在兜圈子，可以直接刪掉，這時候更新答案的值同時把這個點的坐標對應的值更新成最新的（因為以后要是再一次經過的話到當前位置的字符串長度肯定小於到更新前位置的字符串的長度）。最后輸出答案即可。
#include <bits/stdc++.h> using namespace std; int main() { int t; cin>>t; map<pair<int,int>,int>m; while(t--) { int n; scanf("%d",&n); int i; m.clear();//不要忘記清空 m[{0,0}]=0; int x=0,y=0; int ans=100000000; int beg=-1,end=-1; char s[200005]; scanf("%s",s); for(i=1;i<=n;i++) { char c=s[i-1]; if(c=='L') { x-=1; } else if(c=='R') { x+=1; } else if(c=='U') { y+=1; } else if(c=='D') { y-=1; } if(m.find({x,y})!=m.end()) { if(ans>i-m[{x,y}]+1) { ans=i-m[{x,y}]+1;//更新答案 beg=m[{x,y}]+1; end=i; m[{x,y}]=i; } else { m[{x,y}]=i;//ans不必更新的話只更新這個點對應的值即可 } } else { m[{x,y}]=i;//之前沒發現的話直接添加進字典 ，經過第i次操作到達(x,y) } } if(ans!=99999999&&beg!=-1)cout<<beg<<' '<<end<<endl; else cout<<-1<<endl; } return 0; }