codeforces 1272B Snow Walking Robot

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codeforces 1272B Snow Walking Robot

原题地址:https://codeforces.com/problemset/problem/1272/B

题目

Recently you have bought a snow walking robot and brought it home. Suppose your home is a cell (0,0) on an infinite grid.

You also have the sequence of instructions of this robot. It is written as the string s consisting of characters ‘L’, ‘R’, ‘U’ and ‘D’. If the robot is in the cell (x,y) right now, he can move to one of the adjacent cells (depending on the current instruction).

If the current instruction is ‘L’, then the robot can move to the left to (x−1,y);
if the current instruction is ‘R’, then the robot can move to the right to (x+1,y);
if the current instruction is ‘U’, then the robot can move to the top to (x,y+1);
if the current instruction is ‘D’, then the robot can move to the bottom to (x,y−1).
You’ve noticed the warning on the last page of the manual: if the robot visits some cell (except (0,0)) twice then it breaks.

So the sequence of instructions is valid if the robot starts in the cell (0,0), performs the given instructions, visits no cell other than (0,0) two or more times and ends the path in the cell (0,0). Also cell (0,0) should be visited at most two times: at the beginning and at the end (if the path is empty then it is visited only once). For example, the following sequences of instructions are considered valid: “UD”, “RL”, “UUURULLDDDDLDDRRUU”, and the following are considered invalid: “U” (the endpoint is not (0,0)) and “UUDD” (the cell (0,1) is visited twice).

The initial sequence of instructions, however, might be not valid. You don’t want your robot to break so you decided to reprogram it in the following way: you will remove some (possibly, all or none) instructions from the initial sequence of instructions, then rearrange the remaining instructions as you wish and turn on your robot to move.

Your task is to remove as few instructions from the initial sequence as possible and rearrange the remaining ones so that the sequence is valid. Report the valid sequence of the maximum length you can obtain.

Note that you can choose any order of remaining instructions (you don’t need to minimize the number of swaps or any other similar metric).

You have to answer q independent test cases.

Input
The first line of the input contains one integer q (1≤q≤2⋅104) — the number of test cases.

The next q lines contain test cases. The i-th test case is given as the string s consisting of at least 1 and no more than 105 characters ‘L’, ‘R’, ‘U’ and ‘D’ — the initial sequence of instructions.

It is guaranteed that the sum of |s| (where |s| is the length of s) does not exceed 105 over all test cases (∑|s|≤105).

Output
For each test case print the answer on it. In the first line print the maximum number of remaining instructions. In the second line print the valid sequence of remaining instructions t the robot has to perform. The moves are performed from left to right in the order of the printed sequence. If there are several answers, you can print any. If the answer is 0, you are allowed to print an empty line (but you can don’t print it).

Example
inputCopy
6
LRU
DURLDRUDRULRDURDDL
LRUDDLRUDRUL
LLLLRRRR
URDUR
LLL
outputCopy
2
LR
14
RUURDDDDLLLUUR
12
ULDDDRRRUULL
2
LR
2
UD
0

Note
There are only two possible answers in the first test case: “LR” and “RL”.

The picture corresponding to the second test case:
在这里插入图片描述

Note that the direction of traverse does not matter
Another correct answer to the third test case: “URDDLLLUURDR”.

题目大意

给出一串指令,最后要回到原地,且同一个地方除起点外不能经过第二次,你可以删除并重组指令顺序,问最多能走几步,并输出满足的一组指令

解析

因为最后要回到起点,所以上下左右四个方向最后要相互抵消,左右上下的指令数分别相对的一组的较小值,注意其中一组上下或左右没有时,另外一组最多只能有一组。最后按照矩形的方式输出指令

AC代码

#include<cstdio>
#include<iostream>
#include<cstring>
#include<string>
#include<algorithm>
using namespace std;
int n,l,r,u,d,i;
char a[100010]; 
int main(){
	string ans;
	scanf("%d",&n);
	while(n--){
		memset(a,0,sizeof(a));
		l=0;
		r=0;
		d=0;
		u=0;
		scanf("%s",a);
		for(auto b:a)//C++11(如果编译器没有修改过语言标准可能不能编译)
			switch(b){
				case 'L':l++;break;
				case 'R':r++;break;
				case 'D':d++;break;
				case 'U':u++;break;
			}
		l=min(l,r);
		d=min(d,u);
		if(l==0||d==0){
			l=min(l,1);
			d=min(d,1);
		}
		ans="";
		for(i=0;i<l;i++)
			ans+='L';
		for(i=0;i<d;i++)
			ans+='D';
		for(i=0;i<l;i++)
			ans+='R';
		for(i=0;i<d;i++)
			ans+='U';
		cout<<ans.length()<<endl<<ans<<endl;
	}
} 
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