C. Vasya and Robot二分
1.题目描述
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:
- U — move from (x,y) to (x,y+1)
- D — move from (x,y)to (x,y−1)
- L — move from (x,y)to (x−1,y)
- R — move from (x,y) to (x+1,y)
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: maxID−minID+1maxID−minID+1, where maxIDmaxID is the maximum index of a changed operation, and minIDminID is the minimum index of a changed operation. For example, if Vasya changes RRRRRRR to RLRRLRL, then the operations with indices 22, 55 and 77 are changed, so the length of changed subsegment is 7−2+1=67−2+1=6. Another example: if Vasya changes DDDD to DDRD, then the length of changed subsegment is 11.
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.
The 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.
Print 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.
5
RURUU
-2 3
3
4
RULR
1 1
0
3
UUU
100 100
-1
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:
机器人行走的每一步先后顺序其实是没有意义的,这也是这道题的关键。
先按照题目中给的路径计算x,y移动的位置,再二分判断修改的地方在哪里。
代码:
#include<iostream>
#include<stdio.h>
#include<vector>
#include<map>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<set>
#include<cmath>
using namespace std;
const int SIZE = ;
char s[SIZE];
int ex,ey;
int xSum[SIZE];
int ySum[SIZE];
int dx[],dy[];
dy['U'] = ;
dy['D'] = -;
dx['L'] = -;
dx['R'] = ;
//判断下一步在哪个区间段内行走
bool judge(int n, int len, int ex,int ey){
for(int i = ; i + len- <= n; ++i){
int curx = xSum[i-] + xSum[n]-xSum[i+len-];
// 去除长度 len长度 后的 x 走到的位置
int cury = ySum[i-] + ySum[n]-ySum[i+len-];
// 去除长度 len 长度后的 y走到的位置
int delta = abs(curx-ex) + abs(cury-ey);
// 距离到达终点还需要多少步
if(delta <= len && (len-delta)% == ) // 到达终点还需要的步数 一定小于 目前可以通过改变方向的那些步数的个数 len
return true; // 并且 因为此时 len两端 必须是改变的(0 或者 1 是特殊情况) len与delta差值 必须为偶数才能到终点
}
return false;
} int main()
{
int n;
scanf("%d\n%s",&n,s+);
scanf("%d%d",&ex,&ey);
//先算出题目给定路径的最后位置
for(int i = ; i <= n; ++i){
xSum[i] = xSum[i-] + dx[s[i]];
ySum[i] = ySum[i-] + dy[s[i]];
}
int lb = ,ub = n;
int ans = -,mid;
while(lb <= ub){
mid = (lb+ub)/;
if(judge(n,mid,ex,ey)){
ans = mid;
ub = mid-;
}
else{
lb = mid+;
}
}
printf("%d\n",ans);
return ;
}
C. Vasya and Robot二分的更多相关文章
- 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) + ...
- CF 1073C Vasya and Robot(二分答案)
C. Vasya and Robot time limit per test 1 second memory limit per test 256 megabytes input standard i ...
- Educational Codeforces Round 53 (Rated for Div. 2) C. Vasya and Robot 【二分 + 尺取】
任意门:http://codeforces.com/contest/1073/problem/C C. Vasya and Robot time limit per test 1 second mem ...
- Codeforces 1073C:Vasya and Robot(二分)
C. Vasya and Robot time limit per test: 1 secondmemory limit per test: 256 megabytesinput: standard ...
- Codeforces 1073C Vasya and Robot 【二分】
<题目链接> 题目大意: 一个机器人从(0,0)出发,输入一段指令字符串,和机器人需要在指定步数后到达的终点,问如果机器人需要在指定步数内到达终点,那么需要对原指令字符串做出怎样的改变,假 ...
- Educational Codeforces Round 53 (Rated for Div. 2) C. Vasya and Robot(二分或者尺取)
题目哦 题意:给出一个序列,序列有四个字母组成,U:y+1,D:y-1 , L:x-1 , R:x+1; 这是规则 . 给出(x,y) 问可不可以经过最小的变化这个序列可以由(0,0) 变到(x, ...
- 【CF1073C】Vasya and Robot(二分,构造)
题意:给定长为n的机器人行走路线,每个字符代表上下左右走,可以更改将一些字符改成另外三个字符,定义花费为更改的下标max-min+1, 问从(0,0)走到(X,Y)的最小花费,无解输出-1 n< ...
- Codeforces Round #281 (Div. 2) C. Vasya and Basketball 二分
C. Vasya and Basketball time limit per test 2 seconds memory limit per test 256 megabytes input stan ...
- codeforces 676C C. Vasya and String(二分)
题目链接: C. Vasya and String time limit per test 1 second memory limit per test 256 megabytes input sta ...
随机推荐
- 第一节:框架前期准备篇之Log4Net日志详解
一. Log4Net简介 Log4net是从Java中的Log4j迁移过来的一个.Net版的开源日志框架,它的功能很强大,可以将日志分为不同的等级,以不同的格式输出到不同的存储介质中,比如:数据库.t ...
- [物理学与PDEs]第2章习题9 粘性流体动能的衰减
设 $\Omega\subset {\bf R}^3$ 为有界域, ${\bf u}$ 为 Navier-Stokes 方程组 (3. 4)-(3. 5) 满足边界条件 (3. 7) 的解, 其中体积 ...
- java(9)类和对象
一.理解什么是类和对象 万事万物皆对象 1.1.属性——对象具有的特征(特点) 1.2.方法——对象可执行的操作(能干什么事) 1.3.对象的定义: 是一个客观存在的,看的见或摸得着的实体,由属性和方 ...
- 循环语句(for,while,do……while),方法概述
循环结构 分类:for,while,do……while (1)for语句 格式: for(初始化表达式:条件表达式:循环后的操作表达式){ 循环体: } 执行流程: a.执行初始化语句 b.执行判断条 ...
- wc 命令详解
1.wc 命令作用 统计文件里面有多少单词,多少行,多少字符. 2.wc 语法 wc [-lwm] 选项与参数:-l :仅列出行:-w :仅列出多少字(英文单字):-m :多少字符: 3.例子 使用w ...
- 34. Find First and Last Position of Element in Sorted Array
1. 原始题目 给定一个按照升序排列的整数数组 nums,和一个目标值 target.找出给定目标值在数组中的开始位置和结束位置. 你的算法时间复杂度必须是 O(log n) 级别. 如果数组中不存在 ...
- 精通ArrayList,关于ArrayList你想知道的一切
目录 精通ArrayList,关于ArrayList你想知道的一切 前言 ArrayList 内部结构,和常用方法实现 实例化方法 添加元素 add()方法 get()方法 移除元素 怎么扩容的 序列 ...
- 在react-native项目中使用iconfont自定义图标库(android)
1. 安装react-native-vector-icons yarn add react-native-vector-icons react-native link 如果没有关联成功的话,可以参考官 ...
- 图片纯前端JS压缩的实现
一.图片上传前端压缩的现实意义 对于大尺寸图片的上传,在前端进行压缩除了省流量外,最大的意义是极大的提高了用户体验. 这种体验包括两方面: 由于上传图片尺寸比较小,因此上传速度会比较快,交互会更加流畅 ...
- css结构选择器组合使用,选择父元素中多个子元素中某一段元素
nth-of-type()和nth-child()写法一样,这里只用nth-of-type()演示,习惯type 直接上代码 /* 从前向后选择,第6个开始 */ li:nth-of-type(n+6 ...