T3 WRONG - Wrong directions
题面翻译
Farmer John刚刚购买了一台新型可编程拖拉机。为了使拖拉机移动，他输入一串长度为N的绳子
（1 <= N <= 100,000）仅由字符F，L和R组成。每个'F'指示拖拉机向前移动一个单元，
并且字符'L'和'R'分别导致左右转90度。拖拉机从原点开始（0,0）
朝北。
通过输入他想要的命令字符串对他的拖拉机进行编程后，FJ记得他输入了一个字符
命令字符串不正确，但他不记得是哪一个！例如，他可能在打算输入'F'或'L'时
string包含字符'R'。请计算拖拉机可能在飞机上的不同位置的数量
最后结果（拖拉机在最终位置所面向的方向无关紧要）。
题目描述
Farmer John has just purchased a fancy new programmable tractor. To make the tractor move, he types in a string of length N
(1 <= N <= 100,000) consisting of only the characters F, L, and R. Each 'F' instructs the tractor to move forward one unit,
and the characters 'L' and 'R' result in left and right turns of 90 degrees, respectively. The tractor starts out at the origin (0,0)
facing north.
After programming his tractor by typing in his intended command string, FJ remembers that he typed exactly one character in
the command string incorrectly, but he can't remember which one!For example, he might have typed 'F' or 'L' when his intended
string contained the character 'R'. Please compute the number of different locations in the plane at which the tractor might
end up as a result (the direction the tractor faces in its final location does not matter).
Input format: * Line 1: Farmer John's intended command string. SAMPLE INPUT: FF INPUT DETAILS: Farmer John wants the tractor to advance forward twice, ideally ending at position (0,2). OUTPUT FORMAT: * Line 1: The number of positions at which the tractor might end up,given that FJ mistypes one of the characters in his command string. SAMPLE OUTPUT 3 OUTPUT DETAILS: There are 4 possible mistyped sequences: FL, FR, LF, an RF. These will land the tractor at (0,1), (0,1), (-1,0), and (1,0) respectively, a total of 3 distinct locations.
解析
由于我们只修改其中一个字符，所以问题就会变的简单许多。为了减少运算量，可以先预处理出从开始点到中间某一个点时的 \(x\)、\(y\) 和方向。同样预处理出从中间某一个点到终点的 \(x\)、\(y\) 和方向，在预处理时就把起点作为 \((0,0)\)。
对于修改第 \(k\) 个字符，可以把整段路径看作 \(1~k-1\)、\(k\)、\(k+1~len\) 三部分，第一部分和第三部分都预处理过了，只用合并就行了。方向的改变对应着坐标的变换，手推一下就行了。在合并的同时记录一下坐标，可以用 map，我用的哈希表，能快一点。
氧气有毒、建议哈希乘数取131
#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int hmod = 999983, hmax = 1e7 + 5, N = 2e5 + 5;
int len;
int way[4][2] = {{0, 1}, {-1, 0}, {0, -1}, {1, 0}};
int my[4][2] = {{1, 1}, {-1, 1}, {-1, -1}, {1, -1}};
string s;
long long ans;
struct dirt{ int dir, x, y; }ld[N], rd[N];
struct Hash_Table{
int h[hmod+1], nxt[hmax+1], cnt, vx[hmax+1], vy[hmax+1];
void Hash_in(int x, int y){
int hnum = (__int128)(x*131 + y) % hmod;
for(int i=h[hnum]; i; i=nxt[i])
if(vx[i] == x && vy[i] == y) return;
vx[++cnt] = x, vy[cnt] = y, nxt[cnt] = h[hnum], h[hnum] = cnt;
}
bool Hash_out(int x, int y){
int hnum = (__int128)(x*131 + y) % hmod;
for(int i=h[hnum]; i; i=nxt[i])
if(vx[i] == x && vy[i] == y) return 1;
return 0;
}
}h;
inline void merge(int a, char ch, int b){
int x = ld[a].x, y = ld[a].y, d = ld[a].dir;
if(ch == 'F') x += way[d][0], y += way[d][1];
else d = (d + (ch=='L'?1:-1) + 4) % 4;
if(d == 0) x += rd[b].x, y += rd[b].y;
else if(d == 1) x -= rd[b].y, y += rd[b].x;
else if(d == 2) x -= rd[b].x, y -= rd[b].y;
else x += rd[b].y, y -= rd[b].x;
if(!h.Hash_out(x, y)) ++ans, h.Hash_in(x, y);
}
int main(){
freopen("wrongdir.in", "r", stdin);
freopen("wrongdir.out", "w", stdout);
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
cin>>s; len = s.size();
for(int i=1; i<=len; ++i){
if(s[i-1] == 'F'){
ld[i].dir = ld[i-1].dir;
ld[i].x = ld[i-1].x + way[ld[i].dir][0];
ld[i].y = ld[i-1].y + way[ld[i].dir][1];
} else{
ld[i].dir = (ld[i-1].dir + (s[i-1]=='L'?1:-1) + 4) % 4;
ld[i].x = ld[i-1].x, ld[i].y = ld[i-1].y;
}
}
for(int i=len; i>0; --i){
if(s[i-1] == 'F'){
rd[i].dir = rd[i+1].dir;
rd[i].x = rd[i+1].x;
rd[i].y = rd[i+1].y + 1;
} else{
rd[i].x = rd[i+1].y;
rd[i].y = rd[i+1].x * -1;
if(s[i-1] == 'R') rd[i].dir = (rd[i-1].dir + 3) % 4;
else{
rd[i].x *= -1, rd[i].y *= -1;
rd[i].dir = (rd[i-1].dir + 1) % 4;
}
}
}
for(int i=1; i<=len; ++i){
if(s[i-1] == 'F') merge(i-1, 'L', i+1), merge(i-1, 'R', i+1);
else if(s[i-1] == 'R') merge(i-1, 'F', i+1), merge(i-1, 'L', i+1);
else merge(i-1, 'F', i+1), merge(i-1, 'R', i+1);
} return cout<<ans, 0;
}