hash简单处理存图

简单来说  

一般图都是二维或者三维的

这种建图不是特别方便

所以例如二维  对坐标x y进行hash变化

得到一个结果 z 能代表 x 和 y

用map存下 

以代替建立二维图的效果

 

https://codeforces.com/contest/1296/problem/C

C. Yet Another Walking Robot

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Input

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).

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 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.

Example

input

Copy

4
4
LRUD
4
LURD
5
RRUDU
5
LLDDR

output

Copy

1 2
1 4
3 4
-1


对二维图进行变化ha[x*ha+y]=i;

#include<iostream>
#include<map>
#include<algorithm>
 
 
 
using namespace std;
 
typedef long long ll;
map <ll , ll> m;
const int ha =1e6;
const int inf=0x3f3f3f3f;
int main()
{
    ll t,n,x,y;
    char order;
    cin>>t;
    while(t--){
        int l=-1,r=-1,len=inf;
        x=0,y=0;
        m.clear();
        m[0]=0;
        cin>>n;
        for(int i=1;i<=n;i++){
            cin>>order;
            if(order=='U') y++;
            if(order=='L') x--;
            if(order=='R') x++;
            if(order=='D') y--;
            if(m.find(x+ha*y)!=m.end()){
                int temp=m[x+ha*y];
                if(i-temp<len){
                    l=temp;
                    r=i;
                    len=r-l;
                }
            }
            m[x+ha*y]=i;
        }
        if(l==-1&&r==-1) cout<<-1<<endl;
        else cout<<l+1<<' '<<r<<endl;
    }
    return 0;
}

　　

 

呜呜呜  终于会存这种图了   不容易啊