Educational Codeforces Round 53 (Rated for Div. 2) C Vasya and Robot 二分

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 }