Hdoj1325 Is It A Tree ?, Hdoj1325tree

Hdoj1325 Is It A Tree ?, Hdoj1325tree

Is It A Tree? Question Link

Question:
Multiple groups of test data, each group of data has multiple number pairs, indicating a directed edge (that is, the first number is the parent node of the second number), ending with 0, 0 as a group of test data. The test ends when the input is-1 and-1. Determine from the given information whether it is a tree.
Analysis:
1. There can be only one root node, or an empty tree with no points;
2. Except the root node, each vertex has only one parent node.
3. Because there can only be one parent node, We can merge a valid relationship pair (a Father's Day vertex and a child node pair) into a set, their Parent nodes are marked as the root node of the tree to which they belong, so that we can easily determine whether a ring is formed.
4. Determine whether a ring is formed.
5. There is also a very bad situation. Note that not only does the Input-1 and-1 test end, but the input of both numbers is less than 0.

# Include <iostream> # include <cstdio> # include <string. h> # include <math. h ># include <algorithm> using namespace std; const int N = 10005; int a, B, n, mi, mx, sum, flag, v [N], pre [N]; int find (int x) // find the root node of the tree to which x belongs {int r, I, j; r = x; I = x; while (r! = Pre [r]) r = pre [r]; while (pre [I]! = R) {j = pre [I]; pre [I] = r; I = j;} return pre [x];} int main () {n = 0; while (scanf ("% d", & a, & B )! = EOF) {memset (v, 0, sizeof (v); // point involved in the record for (int I = 1; I <= N; I ++) pre [I] = I; if (a <0 & B <0) break; else if (a = 0 & B = 0) // special processing, empty tree {printf ("Case % d is a tree. \ n ", ++ n); continue;} pre [B] = a; v [a] = 1; v [B] = 1; mi = min (, b); mx = max (a, B); flag = 1; while (scanf ("% d", & a, & B )! = EOF) {if (a = 0 & B = 0) break; if (mx <a | mx <B) mx = max (a, B ); if (mi> a | mi> B) mi = min (a, B); v [a] = 1; v [B] = 1; if (flag = 1) {int fx = find (a); int fy = find (B); if (fy! = B) | (fy = fx) // If a ring is formed or the child node already has a parent node, the edge flag is invalid; else if (fy = B & fy! = Fx) // if the ring is not formed and there is no Father's Day, it is legal to add pre [B] = fx ;}} int sum = 0; if (flag = 1) {for (int I = mi; I <= mx; I ++) // determine whether multiple trees exist {if (v [I] = 1) {int fx = find (I); if (fx = I) {sum ++; if (sum> 1) {flag = 0; break ;}}} if (flag = 1) printf ("Case % d is a tree. \ n ", ++ n); else if (flag = 0) printf (" Case % d is not a tree. \ n ", ++ n);} return 0 ;}View Code