HDU 1272 The maze of little Nozomi

http://acm.hdu.edu.cn/showproblem.php?pid=1272

The maze of Xiaoxi

Time limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)
Total submission (s): 34035 Accepted Submission (s): 10406

Problem description Last Gardon 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 of which is a list of integer pairs 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 contains only one row for each set of data that is entered. If the maze conforms to Xiaoxi's idea, then output "Yes", otherwise output "No".

Sample INPUT6 8 5 3 5 2 6 45 6 0 08 1 7 3 6 2 8 9 7 57 4 7 8 7 6 0 03 8 6 8 6 45 3 5 6 5 2 0 0-1-1

Sample Outputyesyesno

Do not need to use and check sets, the number of statistical points and the number of lines, if the number of points-the number of lines = 1, then meet

#include <stdio.h>#include<string.h>#include<math.h>#include<stdlib.h>#defineN 100005intMain () {intA, B, K, M, I; BOOLVis[n];  while(~SCANF ("%d%d", &a, &b)) {k= m =0; memset (Vis,false,sizeof(VIS)); if(A = =-1&& b = =-1)             Break; if(A = =0&& b = =0) {printf ("yes\n"); Continue; }        Else{Vis[a]= Vis[b] =true; K++; }         while(~SCANF ("%d%d", &a, &b)) {if(A = =0&& b = =0)                 Break; Vis[a]= Vis[b] =true; K++; }         for(i =0; i < N; i++)            if(Vis[i]) m++; if(M-k = =1) printf ("yes\n"); Elseprintf ("no\n"); }    return 0;}

HDU 1272 The maze of little Nozomi