CF 1073C Vasya and Robot（二分答案）

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

一个机器人从(0,0)出发，输入一段指令字符串，和机器人需要在指定步数后到达的终点，问如果机器人需要在指定步数内到达终点，那么需要对原指令字符串做出怎样的改变，假设改变 字符串的最大下标为maxindex,改变字符串的最小下标为minindex，输出最小的 maxindex-minindex+1，即，输出最小的改变字符串的连续区间长度(该区间内的字符不一定要全部发生改变)

【分析】

首先考虑在什么情况下，无论如何改动这个字符串都不能到达指定位置

1、字符串长度小于从原点到指定位置的距离

2、字符串长度与从原点到指定位置的奇偶性不同

在除去这两种情况下，剩余的情况都一定有答案。鉴于其可能解时连续的整数，因此，可以用二分枚举所有可能，进而找出最小的连续区间长度。

应注意，当根据给定字符串移动就能到达指定位置，即最小区间为0时，应排除在二分枚举的情况之外。

实际写代码时，特殊情况可以被包含于普通情况。但可以作为思路的引子。

当枚举长度为 x 时，考虑在 string 中所有长度为 x 的子串，是否存在一个子串可行。若存在，尝试缩短子串长度；若不存在，延长子串长度。

判断子串是否可行的方法：

设全集为给定字符串，沿着子串的补集移动，记这样移动到的点为 pos 。求 pos 到 指定位置 的距离，记为 d ，记子串的长度为 len。满足如下两种情况，则子串可行。

1、d <= len

2、(len-d)%2==0

【代码】

#include<cstdio>
#include<cstdlib>
using namespace std;
const int N=2e5+5;
int n,ex,ey,sx[N],sy[N];char s[N];
inline bool check(int m){//假定最佳区间长度为m
	for(int i=1;i+m-1<=n;i++){
		int decx=sx[n]-sx[i+m-1]+sx[i-1];
		int decy=sy[n]-sy[i+m-1]+sy[i-1];
		//不需要改变的区间恒存在的贡献
		int nedx=ex-decx;
		int nedy=ey-decy;
		//需要改变的区间中，x和y想要到达终点，所需恰好作出的贡献
		if(abs(nedx)+abs(nedy)<=m&&!(m-abs(nedx)-abs(nedy)&1)) return 1;
		//(abs(tx)+abs(ty)位字符做出使该人刚好到达终点的贡献，
		//剩下位的字符如果是偶数，就可以让其多走的路程两两抵消，从而刚好到达终点
	}
	return 0;
}
int main(){
	scanf("%d%s%d%d",&n,s+1,&ex,&ey);
	for(int i=1;i<=n;i++){
		sx[i]=sx[i-1]+(s[i]=='L'?-1:(s[i]=='R'?1:0));
		sy[i]=sy[i-1]+(s[i]=='D'?-1:(s[i]=='U'?1:0));
	}
	int l=0,r=n,mid,ans=-1;
	while(l<=r){
		mid=l+r>>1;
		if(check(mid)){
			ans=mid;
			r=mid-1;
		}
		else{
			l=mid+1;
		}
	}
	printf("%d\n",ans);
	return 0;
}