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原题：
C. Vasya and Robot
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Vasya has got a robot which is situated on an infinite Cartesian plane, initially in the cell (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 n operations. Vasya wants to modify this sequence so after performing it the robot will end up in (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+1, where maxID is the maximum index of a changed operation, and minID is the minimum index of a changed operation. For example, if Vasya changes RRRRRRR to RLRRLRL, then the operations with indices 2, 5 and 7 are changed, so the length of changed subsegment is 7−2+1=6. Another example: if Vasya changes DDDD to DDRD, then the length of changed subsegment is 1.
If there are no changes, then the length of changed subsegment is 0. 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) to (x,y), or tell him that it’s impossible.
Input
The first line contains one integer number n (1≤n≤2⋅105) — the number of operations.
The second line contains the sequence of operations — a string of n characters. Each character is either U, D, L or R.
The third line contains two integers 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) to (x,y). If this change is impossible, print −1.
Examples
input
5
RURUU
-2 3
output
3
input
4
RULR
1 1
output
0
input
3
UUU
100 100
output
-1
Note
In the first example the sequence can be changed to LULUU. So the length of the changed subsegment is 3−1+1=3.
In the second example the given sequence already leads the robot to (x,y), so the length of the changed subsegment is 0.
In the third example the robot can’t end his path in the cell (x,y).
中文：
给你一个一个操作字符串，包含UDLR，分别表示为向上向下向左向右移动，然后给你一个坐标(x,y)，让你去修改一段区间的字符串，使得能从原点走到坐标(x,y)，输出这个区间的最小长度，如果不能修改输出-1
代码：
#include<bits/stdc++.h>
using namespace std;typedef long long ll;
const int maxn=200001;int x[maxn],y[maxn],n,X,Y;
string s;int main()
{ios::sync_with_stdio(false);while(cin>>n){cin>>s;cin>>X>>Y;if(abs(X)+abs(Y)>n||(abs(X)+abs(Y)-n)%2){cout<<-1<<endl;continue;}memset(x,0,sizeof(x));memset(y,0,sizeof(y));for(int i=1;i<=n;i++){if(s[i-1]=='U'){y[i]=y[i-1]+1;x[i]=x[i-1];}if(s[i-1]=='D'){y[i]=y[i-1]-1;x[i]=x[i-1];}if(s[i-1]=='R'){x[i]=x[i-1]+1;y[i]=y[i-1];}if(s[i-1]=='L'){x[i]=x[i-1]-1;y[i]=y[i-1];}}int ans=n;int L=0,R=n,tx,ty;int m;while(L<=R){m=(L+R)/2;for(int i=1;i<=n-m+1;i++){tx=x[n]-x[i+m-1]+x[i-1];ty=y[n]-y[i+m-1]+y[i-1];if(abs(X-tx)+abs(Y-ty)>m||(m-abs(X-tx)-abs(Y-ty))%2){if(n-m+1==i)L=m+1;}else{R=m-1;ans=min(ans,m);break;}}}cout<<ans<<endl;}return 0;
}解答：
如果要到达(x,y)，那么只需要横向移动x，纵向移动y即可。所以，如果枚举修改的区间，那么要知道非修改部分有多少横向移动和纵向移动的值。使用预处理前缀和x[i]和y[i]来记录前i个字符串当中横向移动了多少，和纵向移动了多少步。
最后，使用二分或者尺取法枚举区间大小，由于修改区间的中的操作可以变成任意操作，所以只要非区间部分的横向，纵向移动步数和区间长度的关系即可。如果非区间部分的操作在横向和纵向分别已经走了tx和ty，那么横向和纵向剩余的步数abs(x-tx)与abs(x-ty)起码要小于等于修改区间的长度才能实现，此外，如果修改区间的长度减去abs(x-tx)与abs(x-ty)是奇数，那么会多余一个操作或者缺一个操作，使得无法达到(x,y)，根据这两个条件来对区间进行二分即可。
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