HDU 1272 Little Nozomi's Maze (and check)

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 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 Outputyesyesnoproblem-Solving ideas: the title meaning is to find out whether the connection is not linked to the diagram, the first thought is and check set.

1. When judging the ring, just judge the two points of the input side. There is a common parent node, then the two points become rings.

2. When determining connectivity, simply determine the number of root nodes is 1.

Note: When this set of data entered is only 0 0 o'clock, the condition is still satisfied, that is, "Yes" should be output. It is particularly important to note that the border

1#include <iostream>2#include <cstdio>3#include <cstring>4#include <algorithm>5 using namespacestd;6 int Set[100005];7 BOOLvis[100005];8 BOOLFlag;9 intFindintx)Ten { One     intR=x; A      while(Set[r]!=R) -R=Set[R]; -     returnR; the } - intMain () - { -     intb; +     intmax,min; -      while(SCANF ("%d%d", &a,&b), a!=-1&&b!=-1) +     { Aflag=true; at         if(a==0&&b==0) -         { -printf"yes\n"); -             Continue; -         } -          for(intI=1; i<100005; i++) in         { -             Set[i]=i; tovis[i]=false; +         } -          while(a| |b) the         { *vis[a]=true, vis[b]=true; $             intx=find (a);Panax Notoginseng             inty=find (b); -             if(x!=y) the                 Set[x]=y; +             Else Aflag=false; thescanf"%d%d",&a,&b); +         } -         intans=0; $          for(intI=1; i<100005; i++) $         { -             if(Vis[i] &&Set[i]==i) -ans++;//there may be more than one set of topics . the             if(ans>1) -flag=false;Wuyi         } the         if(flag) -printf"yes\n"); Wu         Else -printf"no\n"); About     } $     return 0; -}

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HDU 1272 Little Nozomi's Maze (and check)