C. Vasya and Robot二分

1.题目描述

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 $\mathrm{nn operations.\; Vasya\; wants\; to\; modify\; this\; sequence\; so\; after\; performing\; it\; the\; robot\; will\; end\; up\; in (x,y)(x,y).}$

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

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

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

Input

The first line contains one integer number $\mathrm{n (1\le n\le 2\cdot 105)n (1\le n\le 2\cdot 105) —\; the\; number\; of\; operations.}$

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

The third line contains two integers $\mathrm{x,y (-109\le x,y\le 109)x,y (-109\le x,y\le 109) —\; the\; coordinates\; of\; the\; cell\; where\; the\; robot\; should\; end\; its\; path.}$

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 $\mathrm{3-1+1=33-1+1=3.}$

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 $(x,y)(x,y).$

$2.思路：$

机器人行走的每一步先后顺序其实是没有意义的，这也是这道题的关键。

先按照题目中给的路径计算x,y移动的位置，再二分判断修改的地方在哪里。

代码：

#include<iostream>
#include<stdio.h>
#include<vector>
#include<map>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<set>
#include<cmath>
using namespace std;
const int SIZE = 200005;
char s[SIZE];
int ex,ey;
int xSum[SIZE];
int ySum[SIZE];
int dx[200],dy[200];
dy['U'] = 1;
dy['D'] = -1;
dx['L'] = -1;
dx['R'] = 1;
//判断下一步在哪个区间段内行走 
bool judge(int n, int len, int ex,int ey){
    for(int i = 1; i + len-1 <= n; ++i){
        int curx = xSum[i-1] + xSum[n]-xSum[i+len-1]; 
        // 去除长度 len长度 后的 x 走到的位置
        int cury = ySum[i-1] + ySum[n]-ySum[i+len-1]; 
        // 去除长度 len 长度后的 y走到的位置
        int delta = abs(curx-ex) + abs(cury-ey); 
        // 距离到达终点还需要多少步
        if(delta <= len && (len-delta)%2 == 0) // 到达终点还需要的步数 一定小于 目前可以通过改变方向的那些步数的个数 len
            return true;                       // 并且 因为此时 len两端 必须是改变的(0 或者 1 是特殊情况) len与delta差值 必须为偶数才能到终点
    }
    return false;
}
 
int main()
    {
        int n;
        scanf("%d\n%s",&n,s+1);
        scanf("%d%d",&ex,&ey);
        //先算出题目给定路径的最后位置 
        for(int i = 1; i <= n; ++i){
            xSum[i] = xSum[i-1] + dx[s[i]]; 
            ySum[i] = ySum[i-1] + dy[s[i]];
        }
        int lb = 0,ub = n;
        int ans = -1,mid;
        while(lb <= ub){
            mid = (lb+ub)/2;
            if(judge(n,mid,ex,ey)){
                ans = mid;
                ub = mid-1;
            }
            else{
                lb = mid+1;
            }
        }
        printf("%d\n",ans);
        return 0;
    }