// 403. Frog Jump #include #include #include /* * a frog is crossing a river. the river is divided into some number of units * and at each unit, there may or may not exist a stone. the frog can jump on a * stone, but it must not jump into the water. given a list of 'stones' * positions in (units) in sorted ascending order, determine if the frog can * cross the river by landing on the last stone. initially, the frog is on the * first stone and assumes the frist just must be 1 unit. if the frog's last * jump was k units, its next jujp must be either 'k - 1' or 'k + 1' units. the * frog can only jump in the forward direction. */ bool dfs(int *stones, int stones_size, int t, long long **hash, int x) { if (stones_size == 1) return true; int p = -1, c = -1, n = -1; long long val = (stones[0] << 10) + t; long long d = val; while (1) { if (!hash[d % x]) break; else if (hash[d % x][0] == val) return false; else d++; } for (int i = 1; i < stones_size; i++) { if (stones[i] == stones[0] + t + 1) n = i; if (stones[i] == stones[0] + t) c = i; if (stones[i] == stones[0] + t - 1) p = i; if (stones[i] > stones[0] + t + 1) break; } if (n != -1 && dfs(stones + n, stones_size - n, t + 1, hash, x)) return true; if (c != -1 && dfs(stones + c, stones_size - c, t, hash, x)) return true; if (p != -1 && dfs(stones + p, stones_size - p, t - 1, hash, x)) return true; d = val; while (1) { if (!hash[d % x]) { hash[d % x] = malloc(sizeof(long long)); hash[d % x][0] = val; break; } else d++; } return false; } bool canCross(int *stones, int stones_size) { if (stones[0] || stones[1] != 1) return false; int x = 100 * stones_size; long long **hash = calloc(x, sizeof(long long *)); bool ans = dfs(stones + 1, stones_size - 1, 1, hash, x); for (int i = 0; i < x; i++) { if (!hash[i]) continue; free(hash[i]); } free(hash); return ans; } int main() { int s1[] = {0, 1, 3, 5, 6, 8, 12, 17}, s2[] = {0, 1, 2, 3, 4, 8, 9, 11}; printf("%d\n", canCross(s1, 8)); // expect: 1 printf("%d\n", canCross(s2, 8)); // expect: 0 }