// 2316. Count Unreachable Pairs of Nodes in an Undirected Graph #include #include /* * given integer 'n', there is undirected graph with 'n' nodes, numbered 0 to 'n * - 1' given 2d integer array 'edges' where 'edges[i] = [ai bi]' denotes that * there exists an undirected edge connecting nodes ai and bi. return number of * pairs of different nodes that are unreachable from each other */ int union_find(int *u, int val) { if (u[val] == val) return val; else return union_find(u, u[val]); } void union_set(int *u, int *u_size, int a, int b) { a = union_find(u, a); b = union_find(u, b); if (a == b) return; if (u_size[a] > u_size[b]) { u[b] = u[a]; u_size[a] += u_size[b]; } else { u[a] = u[b]; u_size[b] += u_size[a]; } } long long countPairs(int n, int **edges, int edges_size, int *edges_col_size) { int *u = malloc(n * sizeof(int)), *u_size = malloc(n * sizeof(int)); for (int i = 0; i < n; i++) { u[i] = i; u_size[i] = 1; } for (int i = 0; i < edges_size; i++) union_set(u, u_size, edges[i][0], edges[i][1]); long long ans = 0; for (int i = 0; i < n; i++) ans += n - u_size[union_find(u, i)]; return ans / 2; } int main() { int n1 = 3, e1[][3] = {{0, 1}, {0, 2}, {1, 2}}, es1 = 3, ecs1[] = {3}; int n2 = 7, e2[][5] = {{0, 2}, {0, 5}, {2, 4}, {1, 6}, {5, 4}}, es2 = 5, ecs2[] = {5}; printf("%lld\n", countPairs(n1, e1, es1, ecs1)); // expect: 0 printf("%lld\n", countPairs(n2, e2, es2, ecs2)); // expect: 0 }