D. Buggy Rover
The International Center for Planetary Cartography (ICPC) uses rovers to explore the surfaces of other planets. As we all know, other planets are flat surfaces which can be perfectly and evenly discretized into a rectangular grid structure. Each cell in thi...
Problem Statement
Formatted from the contest statement text, with sample tests broken out into copyable blocks.
rovers to explore the surfaces of other planets. As we all know, other planets are flat surfaces which can be perfectly and evenly discretized into a rectangular grid structure. Each cell in this grid is either flat and can be explored by the rover, or rocky and cannot. Today marks the launch of their brand-new Hornet rover. The rover is set to explore the planet using a simple algorithm. Inter- nally, the rover maintains a direction ordering, a permutation of Mars rover being tested near the Paranal Observatory. the directions north, east, south, and west. When the rover makes CC BY-SA 4.0 by ESO/G. Hudepohl on Wikimedia Commons a move, it goes through its direction ordering, chooses the first direction that does not move it off the face of the planet or onto an impassable rock, and makes one step in that direction. Between two consecutive moves, the rover may be hit by a cosmic ray, replacing its direction ordering with a different one. ICPC scientists have a log of the rover’s moves, but it is difficult to determine by hand if and when the rover’s direction ordering changed. Given the moves that the rover has made, what is the smallest number of times that it could have been hit by cosmic rays?
Input
The first line of input contains two integers r and c, where r (1 ≤ r ≤ 200) is the number of rows on the planet, and c (1 ≤ c ≤ 200) is the number of columns. The rows run north to south, while the columns run west to east. The next r lines each contain c characters, representing the layout of the planet. Each character is either ‘#’, a rocky space; ‘.’, a flat space; or ‘S’, a flat space that marks the starting position of the rover. There is exactly one ‘S’ in the grid. The following line contains a string s, where each character of s is ‘N’, ‘E’, ‘S’, or ‘W’, representing the sequence of the moves performed by the rover. The string s contains between 1 and 10 000 characters, inclusive. All of the moves lead to flat spaces.
Output
Output the minimum number of times the rover’s direction ordering could have changed to be consistent with the moves it made.
Sample Tests
5 3
#..
...
...
...
.S.
NNEN1The rover’s direction ordering could be as follows. In the first move, it either prefers to go north, or it prefers to go south and then north. Note that in the latter case, it cannot move south as it would fall from the face of the planet. In the second move, it must prefer to go north. In the third move, it must prefer to go east. In the fourth move, it can either prefer to go north, or east and then north. It is therefore possible that it was hit by exactly one cosmic ray between the second and third move, changing its direction ordering from N??? to EN?? where ‘?’ stands for any remaining direction.
3 5
.###.
....#
.S...
NEESNS0It is possible the rover began with the direction ordering NESW, which is consistent with all moves it makes.
3 3
...
...
S#.
NEESNNWWSENESS4Editorial
The solution write-up is rendered from the LaTeX source, with equations kept live through MathJax.
Key Observations
There are only $24$ possible direction orderings, so we can treat each ordering as a small DP state.
For a fixed move step and a fixed ordering, it is easy to test whether that ordering is compatible with the recorded move: the chosen direction must be valid, and every valid direction that appears earlier in the ordering must be absent.
Once we know, for every step, which of the $24$ orderings are allowed, the rest is a shortest-path problem on a layered graph: staying with the same ordering costs $0$, switching to a different ordering costs $1$.
Algorithm
Enumerate all $24$ permutations of $\{N,E,S,W\}$.
Simulate the recorded walk once to recover the rover position before each move.
For every step and every permutation, test whether that permutation could have produced the recorded move from that position, and store the result in a boolean table
allowed[step][perm].Run dynamic programming over the move sequence: \[ dp[i][p] = \text{minimum number of changes after processing moves }0..i \] with permutation $p$ used at move $i$. The transition is \[ dp[i][p] = \min_q \bigl(dp[i-1][q] + [p \ne q]\bigr), \] restricted to allowed states.
Correctness Proof
We prove that the algorithm returns the correct answer.
Lemma 1.
For any step and any permutation, allowed[step][perm] is true exactly when that permutation could have produced the recorded move at that step.
Proof.
The rover chooses the first direction in its ordering that leads to a valid neighboring cell. Therefore a permutation is compatible exactly when the recorded direction is valid and every other valid direction that appears earlier in the permutation is impossible. This is exactly the test performed by the program. □
Lemma 2.
The DP value $dp[i][p]$ equals the minimum possible number of ordering changes among all explanations of the first $i+1$ moves that use permutation $p$ on move $i$.
Proof.
The base case $i=0$ is correct because there is no earlier move, so every allowed first permutation costs $0$ changes. For $i>0$, any valid explanation ending with permutation $p$ on move $i$ must come from some permutation $q$ used on move $i-1$. If $q=p$ we pay no extra cost; otherwise we pay exactly one change. Taking the minimum over all valid $q$ gives precisely the recurrence above. □
Theorem.
The algorithm outputs the correct answer for every valid input.
Proof.
By Lemma 1, the DP only considers permutations that are locally consistent with the recorded moves. By Lemma 2, each DP value is the minimum number of changes among all globally consistent explanations with the specified final permutation. Therefore the minimum over the last layer is exactly the minimum number of times the rover's direction ordering could have changed. □
Complexity Analysis
Let $m=|s|$. Testing all step-permutation pairs costs $O(24 \cdot 4 \cdot m)=O(m)$. The DP costs $O(24^2 m)$, which is well within the limits. The memory usage is $O(24m)$ for the compatibility table and $O(24)$ for the rolling DP arrays.
Implementation Notes
The rover position must be updated using the recorded move sequence, not by replaying guessed permutations.
A two-row rolling DP is enough because the transition only depends on the previous step.
Code
C++ solution used for this page.
#include <bits/stdc++.h>
using namespace std;
namespace {
const int INF = (int)1e9;
const int DR[4] = {-1, 0, 1, 0};
const int DC[4] = {0, 1, 0, -1};
int dir_id(char c) {
if (c == 'N') return 0;
if (c == 'E') return 1;
if (c == 'S') return 2;
return 3;
}
} // namespace
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int r, c;
cin >> r >> c;
vector<string> board(r);
for (int i = 0; i < r; ++i) {
cin >> board[i];
}
string path;
cin >> path;
int sr = -1, sc = -1;
for (int i = 0; i < r; ++i) {
for (int j = 0; j < c; ++j) {
if (board[i][j] == 'S') {
sr = i;
sc = j;
}
}
}
vector<array<int, 4>> perm;
array<int, 4> p = {0, 1, 2, 3};
do {
perm.push_back(p);
} while (next_permutation(p.begin(), p.end()));
int P = (int)perm.size();
vector<array<int, 4>> rank(P);
for (int id = 0; id < P; ++id) {
for (int pos = 0; pos < 4; ++pos) {
rank[id][perm[id][pos]] = pos;
}
}
int m = (int)path.size();
vector<vector<char>> good(m, vector<char>(P, 0));
int cr = sr, cc = sc;
for (int step = 0; step < m; ++step) {
int want = dir_id(path[step]);
bool can[4];
for (int d = 0; d < 4; ++d) {
int nr = cr + DR[d];
int nc = cc + DC[d];
can[d] = (0 <= nr && nr < r && 0 <= nc && nc < c && board[nr][nc] != '#');
}
for (int id = 0; id < P; ++id) {
if (!can[want]) {
continue;
}
bool ok = true;
for (int d = 0; d < 4; ++d) {
if (d == want || !can[d]) {
continue;
}
if (rank[id][d] < rank[id][want]) {
ok = false;
break;
}
}
good[step][id] = ok;
}
cr += DR[want];
cc += DC[want];
}
vector<int> dp(P, INF), ndp(P, INF);
for (int id = 0; id < P; ++id) {
if (good[0][id]) {
dp[id] = 0;
}
}
for (int step = 1; step < m; ++step) {
fill(ndp.begin(), ndp.end(), INF);
for (int to = 0; to < P; ++to) {
if (!good[step][to]) {
continue;
}
for (int from = 0; from < P; ++from) {
if (dp[from] == INF) {
continue;
}
ndp[to] = min(ndp[to], dp[from] + (from != to));
}
}
dp.swap(ndp);
}
cout << *min_element(dp.begin(), dp.end()) << '\n';
return 0;
}
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