Codeforces Round #753 (Div. 3)E. Robot on the Board 1

问题翻译：

The robot is located on a checkered rectangular（直角的） board of size n×m (n rows, m columns). The rows in the board are numbered from 1 to n from top to bottom, and the columns — from 1to m from left to right.

有一个方格板n行m列

The robot is able to move from the current cell to one of the four cells adjacent by side.

机器人在方格板上可以从当前位置移动到邻近的时光位置

The sequence of commands s executed by the robot is given. Each command is denoted by one of the symbols 'L', 'R', 'D' or 'U', and triggers the movement to left, right, down or up, respectively.

给出LRDU 指令对应向左右下上移动

The robot can start its movement in any cell. The robot executes the commands starting from the first one, strictly in the order in which they are listed in s. If the robot moves beyond the edge of the board, it falls and breaks. A command that causes the robot to break is not considered successfully executed.

从任意位置开始，依次执行指令

移动到边界，失败

The robot's task is to execute as many commands as possible without falling off the board.

移动次数多且不失败

For example, on board 3×3, if the robot starts a sequence of actions s=s="RRDLUU" ("right", "right", "down", "left", "up", "up") from the central cell, the robot will perform one command, then the next command will force him to cross the edge. If the robot starts moving from the cell (2,1)(2,1) (second row, first column) then all commands will be executed successfully and the robot will stop at the cell (1,2)(1,2) (first row, second column).

The robot starts from cell (2,1) (second row, first column). It moves right, right, down, left, up, and up. In this case it ends in the cell (1,2) (first row, second column).

Determine the cell from which the robot should start its movement in order to execute as many commands as possible.

求出从哪个方格开始移动次数最多

Input

The first line contains an integer tt (1≤t≤1041≤t≤104) — the number of test cases.

The next 2t2t lines contain descriptions of the test cases.

In the description of each test case, the first line contains two integers nn and mm (1≤n,m≤1061≤n,m≤106) — the height and width of the field that the robot is located on. The second line of the description is a string ss consisting solely of characters 'L', 'R', 'D' and 'U' — the sequence of commands the robot executes. The string has a length from 11 to 106106 commands.

It is guaranteed that the total length of ss over all test cases does not exceed 106106.

Output

Print tt lines, each of which contains the answer to the corresponding test case. The answer to the test case are two integers rr (1≤r≤n1≤r≤n) and cc (1≤c≤m1≤c≤m), separated by a space — the coordinates of the cell (row number and column number) from which the robot should start moving to perform as many commands as possible.

If there are several such cells, you may output any of them.

题目概述：

m*n方格给定上下左右走的顺序，走到边界外失败，求在哪个方格走走的步数最多？

思路1（朴素遍历O(N*3）)：

遍历m*n，把每一个方格可以走的步数都求出来，求它们的最大值（test3 超时）

#include<iostream>
#include<vector>
#include<unordered_map>
#include<string>
using namespace std;
bool valid(pair<int, int>p,pair<int,int>b)
{if (p.first < b.first||p.second<b.second||b.first<1||b.second<1){return false;}return true;
}
pair<int, int> board(pair<int, int>p, string s)
{pair<int, int>temp;int count = 0;int maxcount = 0;pair<int, int>maxindex;for (int i = 1; i <= p.first; i++){for (int j = 1; j <= p.second; j++){count = 0;temp.first = i;temp.second = j;pair<int, int>start=temp;for (int k = 0; k < s.size(); k++){if (s[k] == 'L'){temp.second--;if (valid(p, temp)){count++;}else{cout << "坐标" << i << j << "count" << count<<endl;if (maxcount <= count){maxcount = count;maxindex = start;}break;}}else if (s[k] == 'R'){temp.second++;if (valid(p, temp)){count++;}else{cout << "坐标" << i << j << "count" << count<<endl;if (maxcount <= count){maxcount = count;maxindex = start;}break;}}else if (s[k] == 'U'){temp.first--;if (valid(p, temp)){count++;}else{cout << "坐标" << i << j << "count" << count<<endl;if (maxcount <= count){maxcount = count;maxindex = start;}break;}}else if (s[k] =='D'){temp.first++;if (valid(p, temp)){count++;}else{cout << "坐标" << i << j << "count" << count<<endl;if (maxcount <= count){maxcount = count;maxindex = start;}break;}}if (k == s.size() - 1){return start;}}}}return maxindex;}
int main()
{long long m;cin >> m;vector<string>s;string s1;vector<pair<int, int>>vec;pair<int, int>temp;int row;int line;while (m--){cin >> row;cin >> line;temp.first = row;temp.second = line;vec.push_back(temp);cin >> s1;s.push_back(s1);}for (int i = 0; i < vec.size(); i++){pair<int, int>res;res = board(vec[i], s[i]);cout << res.first << " " << res.second << endl;}
}

思路2（朴素遍历的优化（O（n*2）））：（超时）

上下移动和左右移动是两个垂直方向的移动互不影响，所以只要找出行移动最多的位置和列移动最多的位置，就可以唯一确定（或者多个最大值）起始坐标。

统一方法求行和列

时间复杂度降一个数量级

#include<iostream>
#include<vector>
#include<unordered_map>
#include<algorithm>
#include<string>
using namespace std;
int maxx(int i, int size, string s)
{int res = 0;int temp=0;for (int j = 0; j < s.size(); j++){if (s[j] == 'R' || s[j] == 'D'){i++;if (i >= 1 && i <= size){temp++;}else{if (res <= temp){res = temp;}break;}}else if(s[j]=='L'||s[j]=='U'){i--;if (i >= 1 && i <= size){temp++;}else{if (res <= temp){res = temp;}break;}}res = max(temp, res);}return res;
}
int MAX(int size, string s)
{int res=0;int maxindex=1;//从1遍历到size，求起始点最大值for (int i = 1; i <= size; i++){int mm = maxx(i, size, s);if (res <=mm){res = mm;maxindex = i;}}return maxindex;
}
pair<int, int> board(pair<int, int>p, string s)
{string s1;//s1只储存左右方向的移动string s2;//s2只储存上下方向的移动for (int i = 0; i < s.size(); i++){if (s[i] == 'R' || s[i] == 'L'){s1 += s[i];}else{s2 += s[i];}}int m = MAX(p.first, s2);int n = MAX(p.second, s1);pair<int, int>p1;p1.first = m;p1.second = n;return p1;
}
int main()
{long long m;cin >> m;vector<string>s;string s1;vector<pair<int, int>>vec;pair<int, int>temp;int row;int line;while (m--){cin >> row;cin >> line;temp.first = row;temp.second = line;vec.push_back(temp);cin >> s1;s.push_back(s1);}for (int i = 0; i < vec.size(); i++){pair<int, int>res;res = board(vec[i], s[i]);cout << res.first << " " << res.second << endl;}
}

官方思路：（模拟）

翻译：

设棋盘n行*m列

假设点为（r，c），

如果有指令让机器人出界，要么向左一共移动超过r个格子，要么向右一共移动超过m-c+1个格子，要么向上移动超过c个格子，要么向下移动超过n-r+1个格子

失败的思路：

记录四个方向移动的最大值