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：

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <string>
 #include <vector>
 #include <set>
 #include <map>
 #include <stack>
 #include <queue>
 #include <list>
 #include <cstdlib>
 #include <cmath>

 #define FRER() freopen("in.txt", "r", stdin)
 #define FREO() freopen("out.txt", "w", stdout)
 #define INF 0x3f3f3f3f

 using namespace std;

 const int maxn =  + ;

 int n, cx[maxn], cy[maxn], x, y;
 char str[maxn];

 bool judge(int t) {
     for(int i = t; i <= n; ++i) {
         int lx = cx[n] - cx[i] + cx[i - t];
         int ly = cy[n] - cy[i] + cy[i - t];
         if (abs(x - lx) + abs(y - ly) <= t)
             return true;
     }
     return false;
 }

 int main()
 {
     cin >> n >> str >> x >> y;
     if(abs(x) + abs(y) > n || (abs(x) + abs(y)) %  != n % ) {
         cout << - << endl;
     }
     else {
         for(int i = ; i < n; ++i) {
             if(str[i] == 'U') cx[i + ] = cx[i], cy[i + ] = cy[i] + ;
             else if(str[i] == 'D') cx[i + ] = cx[i], cy[i + ] = cy[i] - ;
             else if(str[i] == 'L') cx[i + ] = cx[i] - , cy[i + ] = cy[i];
             else if(str[i] == 'R') cx[i + ] = cx[i] + , cy[i + ] = cy[i];
         }
         int l = , r = n;
         while(l < r) {
             int m = (l + r) >> ;
             if(judge(m))
                 r = m;
             else
                 l = m + ;
         }
         cout << r << endl;
     }
     return ;
 }