题目：题目链接
思路：对于x方向距离与y方向距离之和大于n的情况是肯定不能到达的，另外，如果n比abs(x) + abs(y)大，那么我们总可以用UD或者LR来抵消多余的大小，所以只要abs(x) + abs(y) <= n && (n - abs(x) + abs(y)) % 2 == 0，就一定可以到达终点，判断完之后二分答案长度就可以了
AC代码：
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #include <string> 6 #include <vector> 7 #include <set> 8 #include <map> 9 #include <stack> 10 #include <queue> 11 #include <list> 12 #include <cstdlib> 13 #include <cmath> 14 15 #define FRER() freopen("in.txt", "r", stdin) 16 #define FREO() freopen("out.txt", "w", stdout) 17 #define INF 0x3f3f3f3f 18 19 using namespace std; 20 21 const int maxn = 200000 + 5; 22 23 int n, cx[maxn], cy[maxn], x, y; 24 char str[maxn]; 25 26 bool judge(int t) { 27 for(int i = t; i <= n; ++i) { 28 int lx = cx[n] - cx[i] + cx[i - t]; 29 int ly = cy[n] - cy[i] + cy[i - t]; 30 if (abs(x - lx) + abs(y - ly) <= t) 31 return true; 32 } 33 return false; 34 } 35 36 int main() 37 { 38 cin >> n >> str >> x >> y; 39 if(abs(x) + abs(y) > n || (abs(x) + abs(y)) % 2 != n % 2) { 40 cout << -1 << endl; 41 } 42 else { 43 for(int i = 0; i < n; ++i) { 44 if(str[i] == 'U') cx[i + 1] = cx[i], cy[i + 1] = cy[i] + 1; 45 else if(str[i] == 'D') cx[i + 1] = cx[i], cy[i + 1] = cy[i] - 1; 46 else if(str[i] == 'L') cx[i + 1] = cx[i] - 1, cy[i + 1] = cy[i]; 47 else if(str[i] == 'R') cx[i + 1] = cx[i] + 1, cy[i + 1] = cy[i]; 48 } 49 int l = 0, r = n; 50 while(l < r) { 51 int m = (l + r) >> 1; 52 if(judge(m)) 53 r = m; 54 else 55 l = m + 1; 56 } 57 cout << r << endl; 58 } 59 return 0; 60 }