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// 2392. Build a Matrix With Conditions
#include "leetcode.h"
/*
* you are givne a positive integer 'k'. you are also given a 2d integer array
* 'rowConditions' of size 'n' where 'rowConditions[i] = [above_i, below_i]' and
* a 2d integer array 'colConditions' of size 'm' where 'colConditions[i] =
* [left_i, right_i]'. the two arrays contain integers from 1 to 'k'. you have
* to build a 'k * k' matrix that contains each of the numbers from 1 to 'k'
* exactly once. the remaining cells should satisfy the following conditions.
* the number 'above_i' should appear in a row that is strictly above the row at
* which the number 'below_i' appears for all 'i'. the number 'left_i' should
* appear in a column that is strictly left of the column at which the number
* 'right_i' appears for all 'i'. return any matrix that satisfies these
* conditions.
*/
class Solution {
private:
vector<char> colour;
vector<bool> visited;
bool dfs(int v, vector<vector<int>> &graph) {
colour[v] = 1;
for (int i : graph[v]) {
if (!colour[i]) {
if (dfs(i, graph))
return true;
} else if (colour[i] == 1)
return true;
}
colour[v] = 2;
return false;
}
bool find_cycle(vector<vector<int>> &graph, int n) {
colour.assign(n, 0);
for (int i = 0; i < n; i++)
if (!colour[i] && dfs(i, graph))
return true;
return false;
}
void dfs2(int v, vector<vector<int>> &graph, vector<int> &res) {
visited[v] = true;
for (int i : graph[v])
if (!visited[i])
dfs2(i, graph, res);
res.push_back(v);
}
void topo_sort(vector<vector<int>> &graph, int n, vector<int> &res) {
visited.assign(n, false);
res.clear();
for (int i = 0; i < n; ++i)
if (!visited[i])
dfs2(i, graph, res);
reverse(res.begin(), res.end());
}
public:
vector<vector<int>> buildMatrix(int k, vector<vector<int>> &rowConditions,
vector<vector<int>> &colConditions) {
vector<vector<int>> g1(k), g2(k);
for (auto i : rowConditions)
g1[i[0] - 1].push_back(i[1] - 1);
for (auto i : colConditions)
g2[i[0] - 1].push_back(i[1] - 1);
if (find_cycle(g1, k) || find_cycle(g2, k))
return vector<vector<int>>();
vector<int> a1, a2;
topo_sort(g1, k, a1);
topo_sort(g1, k, a2);
vector<vector<int>> ans(k, vector<int>(k));
map<int, int> m;
for (int i = 0; i < k; i++)
m[a2[i]] = i;
for (int i = 0; i < k; i++)
ans[i][m[a1[i]]] = a1[i] + 1;
return ans;
}
};
int main() {
Solution obj;
vvd(int) rc1 = {{1, 2}, {3, 2}}, cc1 = {{2, 1}, {3, 2}};
vvd(int) rc2 = {{1, 2}, {2, 3}, {3, 1}, {2, 3}}, cc2 = {{2, 1}};
vvd(int) bm1 = obj.buildMatrix(3, rc1, cc1),
bm2 = obj.buildMatrix(3, rc2, cc2);
for (int i = 0; i < bm1.size(); i++) {
for (int j = 0; j < bm1[0].size(); j++)
printf("%d ", bm1[i][j]);
printf("\n");
}
printf("\n");
for (int i = 0; i < bm2.size(); i++) {
for (int j = 0; j < bm2[0].size(); j++)
printf("%d ", bm2[i][j]);
printf("\n");
}
printf("\n");
}
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