Educational Codeforces Round 53 (Rated for Div. 2) C. Vasya and Robot 【二分 + 尺取】

任意门：http://codeforces.com/contest/1073/problem/C

C. Vasya and Robot

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Vasya has got a robot which is situated on an infinite Cartesian plane, initially in the cell (0,0)(0,0). Robot can perform the following four kinds of operations:

• U — move from (x,y) to (x,y+1)
• D — move from (x,y)to (x,y−1)
• L — move from (x,y)to (x−1,y)
• R — move from (x,y) to (x+1,y)

Vasya also has got a sequence of nn operations. Vasya wants to modify this sequence so after performing it the robot will end up in (x,y)(x,y).

Vasya wants to change the sequence so the length of changed subsegment is minimum possible. This length can be calculated as follows: maxID−minID+1maxID−minID+1, where maxIDmaxID is the maximum index of a changed operation, and minIDminID is the minimum index of a changed operation. For example, if Vasya changes RRRRRRR to RLRRLRL, then the operations with indices 22, 55 and 77 are changed, so the length of changed subsegment is 7−2+1=67−2+1=6. Another example: if Vasya changes DDDD to DDRD, then the length of changed subsegment is 11.

If there are no changes, then the length of changed subsegment is 00. Changing an operation means replacing it with some operation (possibly the same); Vasya can't insert new operations into the sequence or remove them.

Help Vasya! Tell him the minimum length of subsegment that he needs to change so that the robot will go from (0,0)(0,0) to (x,y)(x,y), or tell him that it's impossible.

Input

The first line contains one integer number n (1≤n≤2⋅105)n (1≤n≤2⋅105) — the number of operations.

The second line contains the sequence of operations — a string of nn characters. Each character is either U, D, L or R.

The third line contains two integers x,y (−109≤x,y≤109)x,y (−109≤x,y≤109) — the coordinates of the cell where the robot should end its path.

Output

Print one integer — the minimum possible length of subsegment that can be changed so the resulting sequence of operations moves the robot from (0,0)(0,0) to (x,y)(x,y). If this change is impossible, print −1−1.

Examples

input

Copy

5
RURUU
-2 3

output

Copy

3

input

Copy

4
RULR
1 1

output

Copy

0

input

Copy

3
UUU
100 100

output

Copy

-1

Note

In the first example the sequence can be changed to LULUU. So the length of the changed subsegment is 3−1+1=33−1+1=3.

In the second example the given sequence already leads the robot to (x,y)(x,y), so the length of the changed subsegment is 00.

In the third example the robot can't end his path in the cell (x,y)(x,y).

题意概括：

输入N个操作，询问修操作是否能到达终点，如果能到达终点输出“修改的区间”；

修改区间的定义：修改的最大坐标操作的坐标 - 修改的最小坐标操作的坐标

【规定起点为 （0，0），输入终点（ x, y )；】

操作（上下左右）：

• U — move from (x,y) to (x,y+1)
• D — move from (x,y)to (x,y−1)
• L — move from (x,y)to (x−1,y)
• R — move from (x,y) to (x+1,y)

解题思路：二分 || 尺取

预处理路径前缀和（压缩路径）sum_x [ ]，sum_y[ ] ;则sum_x[ N ] , sum_y[ N ] 为实际的终点；

输入的终点为 (ex, ey)，假设能修改若干个操作到达输入的终点，那么：

某一段 [ st, ed ] 所走的影响为:

　　　　　　　　　　　　　 X轴方向：xx = ex - ( sum_x[ N ] - sum_x[ ed -1 ] + sum_x[ st - 1 ] )

　　　　　　　　　　　　 　Y轴方向：yy = ey - ( sum_y[ N ] - sum_y[ ed - 1 ] + sum_y[ st - 1 ] )

二分

二分修改区间长度 len ，尺取判断在该长度是否满足修改条件；

①操作所走最大范围不得超过 len ，因为每次操作只是上下左右移动一步

②判断能否完成假设的影响  len - abs(xx) - abs(yy))%2 ?= 0;

　abs（xx）表示的是 x 方向 到达终点 ex 的差值

　abs（yy）表示的是 y 方向 到达终点 ey 的差值

　假如 len > abs(xx)+abs(yy) 说明这段区间有操作是多余的，但是只要剩下的操作数是偶数就可以两两抵消。

尺取：

直接定义一个头指针一个尾指针，暴力一遍，条件判断是要头指针加还是尾指针加，记录最小修改区间。

AC code:

 #include<bits/stdc++.h>
 using namespace std;
 const int N=1e6+;
 int x[N],y[N];
 int sx,sy,n;
 char s[N];
 bool check(int m)
 {
     for(int i=;i<=n-m+;i++)
     {
         int tx=x[n]-x[i+m-]+x[i-];  //当前原来选项造成的横坐标影响
         int ty=y[n]-y[i+m-]+y[i-];  //当前原来选项造成的纵坐标影响
         int ex=sx-tx, ey=sy-ty;        //消除当前影响
         printf("%d %d m: %d\n",ex,ey,m);
         if(m>=(abs(ex)+abs(ey)) && (m-abs(ex)-abs(ey))%==) //可以构造
             return ;
     }
     return ;
 }
 int main()
 {
     //string s;
     while(~scanf("%d",&n))
     {
         scanf("%s",s+);
         x[]=y[]=;
         for(int i=;i<=n;i++)
         {
             x[i]=x[i-];y[i]=y[i-];  //累积前面步数的结果

             if(s[i]=='L') x[i]-=;    //当前步数造成的影响
             else if(s[i]=='R') x[i]+=;
             else if(s[i]=='D') y[i]-=;
             else y[i]+=;
         }
        // printf("%d %d\n",x[n],y[n]);
         scanf("%d %d",&sx,&sy);       //终点
        // printf("HH");
         int l=,r=n,ans=-;
         while(l<=r)                   //二分答案
         {
             printf("l:%d r:%d\n", l, r);
             int mid=(l+r)>>;
             if(check(mid)) ans=mid,r=mid-;
             else l=mid+;
         }
         printf("%d\n",ans);
     }
 }