HDU 1272 The maze of little Nozomi

Problem Description: Last Gardon's Maze Castle little Nozomi played for a long time (see Problem B), now she also want to design a maze let Gardon to go. But she designed the maze of different ideas, first she thought all the channels should be two-way connectivity, that is, if there is a channel connected to room A and B, then can go through it from room A to room B, but also through it from Room B to room A, in order to improve the difficulty, Xiaoxi hope that any two rooms have and only one path can be connected (unless you go back). Xiao-Nozomi now gives you her design to help you determine whether her design is in line with her design ideas. For example, the first two are eligible, but the last one has two methods of reaching 8 from 5.
Input: Inputs contain multiple sets of data, each set of data is a list of integers ending in 0 0, representing the number of two rooms to which a channel is connected. The number of the room is at least 1 and no more than 100000. There is a blank line between each of the two sets of data.
The entire file ends with two-1. Output: For each set of data entered, the outputs include only one row. If the maze conforms to Xiaoxi's idea, then output "Yes", otherwise output "No".  Sample Input:6 8 5 3 5 2 6 4 5 6 0 0 8 1 7 3 6 2 8 9 7 57 4 7 8 7 6 0 0 3 8 6 8 6 45 3 5 6 5 2 0 0-1-1 Sample Output:yesyesno

Test instructions: Each time the input two points represent an edge, ask if all the edges are entered, whether there is only one path between every two points in the graph (the edges of course are both bidirectional).

Then we just need to determine whether the number of input edges +1 equals the number of points in the graph. (Although this problem is in the topic of the collection, but this method is undoubtedly the best, do not need and check the set can be understood as the minimum spanning tree)

#include <stdio.h>#include<string.h>Const intn=100010;intMain () {intA, B, ans, K, c[n], I;  while(SCANF ("%d%d", &a, &b), A! =-1|| b! =-1)    {        if(A = =0&& b = =0)///at this time the tree is an empty tree{printf ("yes\n"); Continue; } memset (c,0,sizeof(c)); Ans= k =0; C[a]= C[b] =1; Ans++;  while(SCANF ("%d%d", &a, &b), A +b) {ans++; C[a]= C[b] =1; }         for(i =1; i < N; i++)            if(C[i]) k++; if(k = = ans+1) printf ("yes\n"); Elseprintf"no\n"); }    return 0;}

HDU 1272 The maze of little Nozomi