A BFS problem which needs maintain two players location info

Jimmy (xiaoke) Shen
2 min readFeb 3, 2024

As we now that BFS can solve shortest path problem. Today I had a nice question in the ABC contest from atcoder. It is the problem D. For this problem, we can use the standard BFS with slighly modification:

maintain two player’s location info.

As we need two player’s info, the time and space complexity will be

O(n⁴)

Problem

D — Synchronized Player

Code in C++

#include<iostream>                                                                      
#include<vector>
#include<cstring>
using namespace std;
const int N = 65;
int mask[N][N][N][N];
typedef pair<int, int> PII;
bool invalid(int i, int j, vector<string>&grid, int n){
if (i == 0 || j ==0 || i == n +1 || j == n +1 || grid[--i][--j] == '#')return true;
return false;
}
int bfs(int i1, int j1, int i2, int j2, vector<string>& grid, int n){
vector<pair<PII, PII>> curr_level;
vector<PII> dirs = {{0, 1},{0, -1},{1, 0},{-1, 0}};
curr_level.emplace_back(make_pair(i1, j1), make_pair(i2, j2));
int level = 0;
int r1, r2, c1, c2;
while (!curr_level.empty()){

vector<pair<PII, PII>> next_level;
for (auto [ij1, ij2]: curr_level){
auto [i1, j1] = ij1;
auto [i2, j2] = ij2;
for (auto [di, dj]: dirs){
r1 = i1 + di;
c1 = j1 + dj;
r2 = i2 + di;
c2 = j2 + dj;
if (invalid(r1, c1, grid, n)){
r1 = i1;
c1 = j1;
}
if (invalid(r2, c2, grid, n)){
r2 = i2;
c2 = j2;
}
if (r1 == r2 && c1 == c2)return level + 1;
if (mask[r1][c1][r2][c2])continue;
mask[r1][c1][r2][c2] = 1;
next_level.emplace_back(make_pair(r1, c1), make_pair(r2, c2));
}
}
level++;
curr_level = next_level;
}
return -1;
}
int main(){
memset(mask, 0, sizeof mask);
int n; cin >> n;
int i1=-1, j1=-1, i2=-1, j2 =-1;
vector<string> grid(n);
for (int i = 0; i < n; ++i){
cin >> grid[i];
for (int j = 0; j < n; ++j){
if (grid[i][j] == 'P'){
if (i1 == -1){
i1 = i + 1;
j1 = j + 1;
}else{
i2 = i + 1;
j2 = j + 1;
}
}
}
}
mask[i1][j1][i2][j2] = 1;
auto ret = bfs(i1, j1, i2, j2, grid, n);
cout << ret << endl;
return 0;
}

The code is very standard BFS. Using this code can solve the problem smoothly!

Happy coding!

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