// 1514. Path with Maximum Probability #include #include #include #include /* * given an undirected weighted graph of 'n' nodes (0-indexed). represented by * an edge list where 'edges[i] = [a, b]' is an undirected edge connecting the * nodes 'a' and 'b' with a probability of success of traversing that edge * 'succ_prob[i]'. given two nodes 'start, end', find the path with the maximum * probability of success to go from 'start' to 'end' and return its success * probability. if there is no path from 'start' to 'end', return 0. your answer * will be accepted if it differs from the correct answer by at most '1e-5' */ #define LOWER_BOUND 1e-5 typedef struct { int next; double val; } node; void dfs(int n, bool *visited, double val, double *ans, int curr, int end, int *cn, node **map) { if (curr == end) *ans = fmax(*ans, val); if (*ans >= val) return; if (val - *ans <= LOWER_BOUND) return; if (visited[curr]) return; visited[curr] = true; for (int i = 0; i < cn[curr]; i++) { double new_val = val * map[curr][i].val; dfs(n, visited, new_val, ans, map[curr][i].next, end, cn, map); } visited[curr] = false; } double maxProbability(int n, int **edges, int edges_size, int *edges_col_size, double *succ_prob, int succ_prob_size, int start, int end) { double ans = 0; bool *visited = calloc(n, sizeof(bool)); int *cn = calloc(n, sizeof(int)); node **map = calloc(n, sizeof(node *)); for (int i = 0; i < edges_size; i++) { cn[edges[i][0]]++; cn[edges[i][1]]++; } for (int i = 0; i < n; i++) if (cn[i]) map[i] = malloc(cn[i] * sizeof(node)); int *m_idx = calloc(n, sizeof(int)); for (int i = 0; i < edges_size; i++) { int prob = edges[i][0], next = edges[i][1], idx = m_idx[prob]; map[prob][idx].next = next; map[prob][idx].val = succ_prob[i]; m_idx[prob]++; prob = edges[i][1]; next = edges[i][0]; idx = m_idx[prob]; map[prob][idx].next = next; map[prob][idx].val = succ_prob[i]; m_idx[prob]++; } dfs(n, visited, 1.0, &ans, start, end, cn, map); free(visited); for (int i = 0; i < n; i++) if (cn[i]) free(map[i]); free(cn); free(map); return ans; }