Hdoj 1272 The maze of little Nozomi

The maze of Xiaoxi

Time limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)
Total submission (s): 30827 Accepted Submission (s): 9540

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 Outputyesyesno

#include <stdio.h>int set[101000];int find (int fa) {int ch=fa;int t;while (Fa!=set[fa]) fa=set[fa];while (CH!=FA) { t=set[ch];set[ch]=fa;ch=t;} return FA;} int mix (int x,int y) {int fx,fy;fx=find (x); Fy=find (y); if (Fx!=fy) set[fx]=fy;} int main () {int i,s;int fang,fang1;//input two point int way;//record how many paths are reached for a node while (1) {for (i=1;i<101000;i++)    set[i]=0; way=0;    while (scanf ("%d%d", &fang,&fang1) && (fang!=0| | fang1!=0)    {    if (fang==-1&&fang1==-1) return 0;if (set[fang]==0) set[fang]=fang;if (set[fang1]==0) Set[fang1]=fang1;if (Find (Fang) ==find (fang1))//If there is a way to make way++ before merging;                      The root node of this point is the same as the root node of the current point, there are multiple paths Mix (fang,fang1);//merge existing Paths    }    if (way!=0)    printf ("no\n");       else    {        s=0;    for (i=1;i<=101000;i++)    {    if (set[i]==i)   //If more than one root node proves that all rooms are not fully connected     s++;    }    if (s>1)    printf ("no\n");  Output note case     else    printf ("yes\n");}    } return 0;}

　　

Hdoj 1272 Little Greek maze