1.題目描述
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:
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
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
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).
2.思路：
機器人行走的每一步先後順序其實是沒有意義的，這也是這道題的關鍵。
先按照題目中給的路徑計算x,y移動的位置，再二分判斷修改的地方在哪裡。
程式碼：
1 #include<iostream> 2 #include<stdio.h> 3 #include<vector> 4 #include<map> 5 #include<cstring> 6 #include<cstdlib> 7 #include<algorithm> 8 #include<set> 9 #include<cmath> 10 using namespace std; 11 const int SIZE = 200005; 12 char s[SIZE]; 13 int ex,ey; 14 int xSum[SIZE]; 15 int ySum[SIZE]; 16 int dx[200],dy[200]; 17 dy['U'] = 1; 18 dy['D'] = -1; 19 dx['L'] = -1; 20 dx['R'] = 1; 21 //判斷下一步在哪個區間段內行走 22 bool judge(int n, int len, int ex,int ey){ 23 for(int i = 1; i + len-1 <= n; ++i){ 24 int curx = xSum[i-1] + xSum[n]-xSum[i+len-1]; 25 // 去除長度 len長度 後的 x 走到的位置 26 int cury = ySum[i-1] + ySum[n]-ySum[i+len-1]; 27 // 去除長度 len 長度後的 y走到的位置 28 int delta = abs(curx-ex) + abs(cury-ey); 29 // 距離到達終點還需要多少步 30 if(delta <= len && (len-delta)%2 == 0) // 到達終點還需要的步數 一定小於 目前可以通過改變方向的那些步數的個數 len 31 return true; // 並且 因為此時 len兩端 必須是改變的(0 或者 1 是特殊情況) len與delta差值 必須為偶數才能到終點 32 } 33 return false; 34 } 35 36 int main() 37 { 38 int n; 39 scanf("%d\n%s",&n,s+1); 40 scanf("%d%d",&ex,&ey); 41 //先算出題目給定路徑的最後位置 42 for(int i = 1; i <= n; ++i){ 43 xSum[i] = xSum[i-1] + dx[s[i]]; 44 ySum[i] = ySum[i-1] + dy[s[i]]; 45 } 46 int lb = 0,ub = n; 47 int ans = -1,mid; 48 while(lb <= ub){ 49 mid = (lb+ub)/2; 50 if(judge(n,mid,ex,ey)){ 51 ans = mid; 52 ub = mid-1; 53 } 54 else{ 55 lb = mid+1; 56 } 57 } 58 printf("%d\n",ans); 59 return 0; 60 }View Code
