HDU 1272 Little Nozomi's Maze (and check)

The maze of XiaoxiTime limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)
Total submission (s): 27851 Accepted Submission (s): 8613


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

First, we learned two points:

Determine whether a graph is the two key points of a tree:

1. No loops exist
2. A rule that satisfies the number of edges plus a vertex (regardless of the heavy edges and the edges pointing to itself)

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.
The number of edges and vertices, in the union when the number of sides plus one, the top points in the reading when the statistics, the final judgment can be, the code is written down, because the back to see others have to blog to know, input will only 0 0 such data, which is a reasonable input, to output Yes

#include <iostream> #include <stdio.h>using namespace std; #define M 100010int pre[m];int cnt[m];int flag=0;    int global_edges=0;int _find (int node) {if (Node==pre[node]) return node; else return Pre[node]=_find (Pre[node]);}    int _union (int a,int b) {if (flag) return 1;    int P=_find (a);    int Q=_find (b);    if (p==q) return 1;        else {pre[a]=b;        global_edges++;    return 0;   }}int Main (int argc, char *argv[]) {int yflag=0;    Freopen ("1272.in", "R", stdin);    int A, B;        for (int i=1;i<m;++i) {pre[i]=i;    cnt[i]=0;        } while (scanf ("%d%d", &a,&b)) {if (a==-1&&b==-1) return 0;            else if (a==0&&b==0) {int sum=0;            for (int i=1;i<m;++i) {sum+=cnt[i];            } if (yflag==0) printf ("yes\n");            else if (flag) printf ("no\n");            else if (sum!=global_edges+1) printf ("no\n"); else printf ("yes\n");            flag=0;            global_edges=0;                for (int i=1;i<m;++i) {pre[i]=i;            cnt[i]=0;        } yflag=0;            } else {yflag=1;            if (a<b) flag=_union (A, b);            else Flag=_union (b,a);            if (cnt[a]==0) cnt[a]++;        if (cnt[b]==0) cnt[b]++; }} return 0;}

HDU 1272 Little Nozomi's Maze (and check)