Hdoj 1272 The maze of little Nozomi

The maze of XiaoxiTime limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others) Total Submission (s): 35613 Accepted Submission (s): 10881

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 Input

6 8  5 3 5 2 6 6 0 1 7 3  6 2  8 9  7  4 7 8 7 6 0 8 6 8 6 4 5 3  5 6  5 2  0 0-1-1

Sample Output

Yesyesno

The subject has a pit, when input 0 0 o'clock should output yes!                       Title requirements: 1. There must be a tree 2. Can't be a ring! 3. And to find the upper and lower limits of the node!

Include<stdio.h>int set[100050];int v[100050];int Find (int p) {int t;    int child=p;    while (P!=set[p])//Find root node p=set[p];        while (child!=p) {T=set[child];        Set[child]=p;    child=t; } return p;}    int merge (int x,int y)//merge {int fx=find (x);    int Fy=find (y);        if (fx!=fy) {set[fx]=fy;    return 1; } else return 0;}    int main () {int a A, B;    int i;        while (scanf ("%d%d", &a,&b)!=eof&&a!=-1&&b!=-1) {int w=1;         if (a==0&&b==0)//This is a big pit!!!!            {printf ("yes\n");        Continue             } for (i=0;i<=100050;i++) {set[i]=i;        Whether the v[i]=0;//tag has been accessed} int min=9999999,max=-1; while (a| |            b)//Find out the maximum and minimum values of {if (A>max) max=a;            if (B>max) max=b;            if (a<min) min=a;        if (b<min) min=b; V[a]=1; The v[b]=1;//is labeled as 1 if (!merge (A, a))//Whether it is a ring, that is, whether it complies with test instructions          w=0;         scanf ("%d%d", &a,&b);         } int count=0;         if (w==0) printf ("no\n");                   else {for (i=min;i<=max;i++)//Find the number of trees {if (v[i]&&set[i]==i)              count++;              } if (count==1) printf ("yes\n");        else printf ("no\n"); }} return 0; }

Original title: Hdoj 1272 Little Greek Maze

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Hdoj 1272 The maze of little Nozomi