What is it a tree in poj 1308?

In addition to the empty tree, the tree has only one root node. Besides the root node, the other nodes have only one precursor.

To determine whether a graph is a tree, you need to determine whether there is only one common ancestor, and the inbound degree is less than 2, and the ring cannot be formed.

Link: poj 1308

# Include <stdio. h> # define n 100000int f [n + 10], t [n + 10], C [n + 10]; int find (int A) {if (! = F [a]) f [a] = find (F [a]); Return f [a];} int mix (int A, int B) {int X, y; X = find (a); y = find (B); If (x = y) return 0; F [x] = y; return 1 ;} int main () {int I, j = 0, n, m, a, B, flag; while (scanf ("% d", & M, & N )! = EOF) {If (M <0 & n <0) // terminate break when two negative numbers are input; j ++; if (M = 0 & n = 0) {printf ("case % d is a tree. \ n ", J); continue;} for (I = 1; I <= N; I ++) {f [I] = I; c [I] = T [I] = 0;} flag = 1; while (M! = 0 & n! = 0) {T [m] = T [N] = 1; C [N] ++; // calculate the inbound degree if (C [N]> 1) flag = 0; A = mix (m, n); // determine whether the ring is formed if (! A) Flag = 0; scanf ("% d", & M, & N) ;}if (FLAG) {B = 0; for (I = 1; I <= N; I ++) {If (T [I] & F [I] = I) B ++; // calculate the number of root nodes if (B> 1) break;} If (flag & B = 1) printf ("case % d is a tree. \ n ", J); else printf (" case % d is not a tree. \ n ", J);} return 0 ;}