Yet Another Walking Robot （Map）

题意描述

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

思路

这道题想优化想了很长时间，于是就去看了题解。原来的思路是寻找给定字符串的子串，字串内的操作能够抵消。而题解的做法是记录坐标，如果该坐标之前已经存在，则找到了答案。自己看问题还是不够全面，看到字符串就向字符串的角度去想，而忽视了其他重要的因素。

AC代码

#include<bits/stdc++.h>
#define x first
#define y second
#pragma GCC optimize(2)
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC optimize(3)
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC target("sse3","sse2","sse")
#pragma GCC target("avx","sse4","sse4.1","sse4.2","ssse3")
#pragma GCC target("f16c")
#pragma GCC optimize("inline","fast-math","unroll-loops","no-stack-protector")
#pragma GCC diagnostic error "-fwhole-program"
#pragma GCC diagnostic error "-fcse-skip-blocks"
#pragma GCC diagnostic error "-funsafe-loop-optimizations"
#pragma GCC diagnostic error "-std=c++14"
#define IOS ios::sync_with_stdio(false);cin.tie(0);
using namespace std;
typedef unsigned long long ULL;
typedef pair<int,int> PII;
typedef long long LL;
const int N=2*1e5+10;
const int M=150;
const int INF=0x3f3f3f3f;
const int MOD=998244353;
map<PII,int> m;
int main(){
    IOS;
    int T;cin>>T;
    while(T--){
        int n;cin>>n;
        string s;cin>>s;
        m.clear();
        PII ans={0,0};
        m[{0,0}]=0;
        int x=0,y=0;
        int cnt=INF;
        for(int i=1;i<=n;i++){
            if(s[i-1]=='L') x--;
            if(s[i-1]=='R') x++;
            if(s[i-1]=='U') y++;
            if(s[i-1]=='D') y--;
            if(m.count({x,y})){
                int a=m[{x,y}];
                if(i-a<cnt){
                    cnt=i-a;
                    ans={a+1,i};
                }
            }
            m[{x,y}]=i;
        }
        if(ans.first==0) cout<<"-1"<<endl;
        else cout<<ans.first<<' '<<ans.second<<endl;

    }
    return 0;
}