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Unique-Paths-III.cpp
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Unique-Paths-III.cpp
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// I am using here backtracking with memoization (look at 'used').
class Solution {
private:
int n;
int m;
int answer = 0;
int required = 0;
bool isValid(int x, int y) {
return (0 <= x && x < n) && (0 <= y && y < m);
}
void dfs(vector<vector<int>> &grid, pair<int, int> s, pair<int, int> e, vector<pair<int, int>>& st, vector<vector<bool>> &used) {
if (!st.empty() && st.back() == e && st.size() == required - 1) {
++answer;
return;
}
vector<int> dx = {0, 1, 0, -1};
vector<int> dy = {-1, 0, 1, 0};
used[s.first][s.second] = true;
for (int i = 0; i < 4; ++i) {
if (isValid(s.first + dx[i], s.second + dy[i]) && (grid[s.first + dx[i]][s.second + dy[i]] == 0 || grid[s.first + dx[i]][s.second + dy[i]] == 2) && !used[s.first + dx[i]][s.second + dy[i]]) {
used[s.first + dx[i]][s.second + dy[i]] = true;
st.push_back({s.first + dx[i], s.second + dy[i]});
dfs(grid, {s.first + dx[i], s.second + dy[i]}, e, st, used);
st.pop_back();
used[s.first + dx[i]][s.second + dy[i]] = false;
}
}
}
public:
int uniquePathsIII(vector<vector<int>>& grid) {
n = (int) grid.size();
m = (int) grid[0].size();
pair<int, int> s;
pair<int, int> e;
for (int i = 0; i < n; ++i) {
for (int j = 0; j < m; ++j) {
if (grid[i][j] == 1) {
s = {i, j};
++required;
}
if (grid[i][j] == 2) {
e = {i, j};
++required;
}
if (grid[i][j] == 0) {
++required;
}
}
}
vector<pair<int, int>> st;
vector<vector<bool>> used(n, vector<bool> (m, false));
used[s.first][s.second] = true;
dfs(grid, s, e, st, used);
return answer;
}
};