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// 1368. Minimum Cost to Make at Least One Valid Path in a Grid
#include "leetcode.h"
/*
* given an 'm x n' grid. each cell of the grid has a sign pointing to the next
* cell you shhould visit if you are currently in this cell. notice that there
* could be some signs on the cells of the grid that point outside the grid. you
* will initially start at the upper left cell '(0, 0)'. a valid path in the
* grid is a path that starts from the uppper left cell '(0, 0)' and ends at the
* bbbottom right cell '(m - 1, n - 1)' following the signs on the grid. the
* valid path does not have to bbe the shortest. return the minimum cost to make
* the grid have at least one valid path
*/
struct node {
int state[2];
int moves;
};
struct queue {
struct node data[10000];
int front, rear;
};
void queue_init(struct queue *obj) {
obj->front = -1;
obj->rear = -1;
}
bool queue_empty(struct queue *obj) {
return obj->front == -1 && obj->rear == -1;
}
void queue_push(struct queue *obj, int *state, int moves) {
if (queue_empty(obj)) {
obj->front = 0;
obj->rear = 0;
} else
obj->rear = (obj->rear + 1) % 10000;
obj->data[obj->rear].state[0] = state[0];
obj->data[obj->rear].state[1] = state[1];
obj->data[obj->rear].moves = moves;
}
struct node queue_pop(struct queue *obj) {
struct node n = obj->data[obj->front];
if (obj->front == obj->rear) {
obj->front = -1;
obj->rear = -1;
} else
obj->front = (obj->front + 1) % 10000;
return n;
}
int minCost(int **grid, int gridSize, int *gridColSize) {
int n = gridSize, m = *gridColSize, min = INT_MAX;
int vis[100][100] = {0};
int **dp = (int **)malloc((n + 1) * sizeof(int *));
for (int i = 0; i < n + 1; i++)
dp[i] = (int *)malloc((m + 1) * sizeof(int));
for (int i = 0; i < n + 1; i++)
for (int j = 0; j < m + 1; j++)
dp[i][j] = -1;
struct queue q;
queue_init(&q);
int init_state[2] = {0, 0};
queue_push(&q, init_state, 0);
dp[0][0] = 0;
while (!queue_empty(&q)) {
struct node curr = queue_pop(&q);
int x = curr.state[0], y = curr.state[1];
int dir[5][2] = {{0, 0}, {0, 1}, {0, -1}, {1, 0}, {-1, 0}};
for (int i = 1; i < 5; i++) {
int curr_move = 0, next[2];
if (grid[x][y] != i)
curr_move = 1;
next[0] = x + dir[i][0];
next[1] = y + dir[i][1];
int next_move = curr.moves + curr_move;
if (next[0] == n && next[1] == m && min < next_move)
min = next_move;
if (next[0] >= 0 && next[1] >= 0 && next[0] < n && next[1] < m &&
(!vis[next[0]][next[1]] || dp[next[0]][next[1]] > next_move)) {
dp[next[0]][next[1]] = next_move;
vis[next[0]][next[1]] = 1;
queue_push(&q, next, next_move);
}
}
}
int ans = dp[n - 1][m - 1];
for (int i = 0; i < n + 1; i++)
free(dp[i]);
free(dp);
return ans;
}
int main() {
int g1i[4][4] = {{1, 1, 1, 1}, {2, 2, 2, 2}, {1, 1, 1, 1}, {2, 2, 2, 2}};
int g2i[3][3] = {{1, 1, 3}, {3, 2, 2}, {1, 1, 4}};
int g3i[2][2] = {{1, 2}, {4, 3}};
struct two_d_arr g1, g2, g3;
two_d_arr_init(&g1, ARRAY_SIZE(g1i), ARRAY_SIZE(g1i[0]), g1i);
two_d_arr_init(&g2, ARRAY_SIZE(g2i), ARRAY_SIZE(g2i[0]), g2i);
two_d_arr_init(&g3, ARRAY_SIZE(g3i), ARRAY_SIZE(g3i[0]), g3i);
printf("%d\n", minCost(g1.arr, g1.row_size, g1.col_size)); // expect: 3
printf("%d\n", minCost(g2.arr, g2.row_size, g2.col_size)); // expect: 0
printf("%d\n", minCost(g3.arr, g3.row_size, g3.col_size)); // expect: 1
two_d_arr_free(&g1);
two_d_arr_free(&g2);
two_d_arr_free(&g3);
}
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