任意門：http://codeforces.com/contest/1073/problem/C
Vasya has got a robot which is situated on an infinite Cartesian plane, initially in the cell
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:
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.
InputThe 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.
OutputPrint 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 Copy5output Copy
RURUU
-2 3
3input Copy
4output Copy
RULR
1 1
0input Copy
3output Copy
UUU
100 100
-1Note
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).
題意概括：
輸入N個操作，詢問修操作是否能到達終點，如果能到達終點輸出“修改的區間”；
修改區間的定義：修改的最大座標操作的座標 - 修改的最小座標操作的座標
【規定起點為 （0，0），輸入終點（ x, y )；】
操作（上下左右）：
解題思路：二分 || 尺取
預處理路徑字首和（壓縮路徑）sum_x [ ]，sum_y[ ] ;則sum_x[ N ] , sum_y[ N ] 為實際的終點；
輸入的終點為 (ex, ey)，假設能修改若干個操作到達輸入的終點，那麼：
某一段 [ st, ed ] 所走的影響為:
X軸方向：xx = ex - ( sum_x[ N ] - sum_x[ ed -1 ] + sum_x[ st - 1 ] )
Y軸方向：yy = ey - ( sum_y[ N ] - sum_y[ ed - 1 ] + sum_y[ st - 1 ] )
二分
二分修改區間長度 len ，尺取判斷在該長度是否滿足修改條件；
①操作所走最大範圍不得超過 len ，因為每次操作只是上下左右移動一步
②判斷能否完成假設的影響 len - abs(xx) - abs(yy))%2 ?= 0;
abs（xx）表示的是 x 方向 到達終點 ex 的差值
abs（yy）表示的是 y 方向 到達終點 ey 的差值
假如 len > abs(xx)+abs(yy) 說明這段區間有操作是多餘的，但是隻要剩下的運算元是偶數就可以兩兩抵消。
尺取：
直接定義一個頭指標一個尾指標，暴力一遍，條件判斷是要頭指標加還是尾指標加，記錄最小修改區間。
AC code:
1 #include<bits/stdc++.h> 2 using namespace std; 3 const int N=1e6+6; 4 int x[N],y[N]; 5 int sx,sy,n; 6 char s[N]; 7 bool check(int m) 8 { 9 for(int i=1;i<=n-m+1;i++) 10 { 11 int tx=x[n]-x[i+m-1]+x[i-1]; //當前原來選項造成的橫座標影響 12 int ty=y[n]-y[i+m-1]+y[i-1]; //當前原來選項造成的縱座標影響 13 int ex=sx-tx, ey=sy-ty; //消除當前影響 14 printf("%d %d m: %d\n",ex,ey,m); 15 if(m>=(abs(ex)+abs(ey)) && (m-abs(ex)-abs(ey))%2==0) //可以構造 16 return 1; 17 } 18 return 0; 19 } 20 int main() 21 { 22 //string s; 23 while(~scanf("%d",&n)) 24 { 25 scanf("%s",s+1); 26 x[0]=y[0]=0; 27 for(int i=1;i<=n;i++) 28 { 29 x[i]=x[i-1];y[i]=y[i-1]; //累積前面步數的結果 30 31 if(s[i]=='L') x[i]-=1; //當前步數造成的影響 32 else if(s[i]=='R') x[i]+=1; 33 else if(s[i]=='D') y[i]-=1; 34 else y[i]+=1; 35 } 36 // printf("%d %d\n",x[n],y[n]); 37 scanf("%d %d",&sx,&sy); //終點 38 // printf("HH"); 39 int l=0,r=n,ans=-1; 40 while(l<=r) //二分答案 41 { 42 printf("l:%d r:%d\n", l, r); 43 int mid=(l+r)>>1; 44 if(check(mid)) ans=mid,r=mid-1; 45 else l=mid+1; 46 } 47 printf("%d\n",ans); 48 } 49 }View Code
