nyoj-42-a stroke problem-dfs

Time limit for a stroke problem: theMs | Memory Limit:65535KB Difficulty:4

Describe

Zyc like to play some small games since childhood, including painting a stroke, he would like to ask you to help him write a program, judge whether a figure can use a stroke down.

Rules, all sides can only draw once, can not repeat the painting.

Input

the first line has only one positive integer N (n<=10) that represents the number of groups of test data.
The first line of each set of test data has two positive integer p,q (p<=1000,q<=2000), representing how many vertices and lines are in the picture. (number of points from 1 to p)
The subsequent Q line has two positive integers, a, a, B (0<a,b<p) for each line, indicating that there is a connection between the two points numbered a and a.

Output

If there is a qualifying connection, output "Yes",
If there are no qualifying lines, output "no".

Sample input

24 31 21 31 44 51 22 31 31 43 4

Sample output

NoYes

#include <cstdio> #include <cstring> #include <cmath> #include <cstdlib> #include <iostream   > #include <algorithm> #include <queue> #include <stack>using namespace Std;int line[1010];   Statistic the degree of each node int visit[1010]; Record the visited point int g[1010][1010]; Record whether i,j is connected to int vertex,edge;    void DFS (int i)//determine connectivity {//int v=i;visit[i]=1; I have visited for (int v=0;v<=vertex;v++) {if (v!=i && g[i][v] &&!visit[v])//Find one! =i/With I connect/no access point DFS (v);//Continue from this point down access}}int main () {int t,x,y,count,flag;while (~scanf ("%d", &t)) {while (t--) {Coun T=0;FLAG=1;SCANF ("%d%d", &vertex,&edge); Number of vertices and number of sides memset (line), memset (visit,0,sizeof (visit)), memset (g,0,sizeof (G)); for (int i=1;i<=edge; i++) {scanf ("%d%d", &x,&y); G[x][y]=g[y][x]=1;    line[x]++;    The degree of the point line[y]++; }dfs (0); for (int i=0;i<=vertex;i++) {g[i][i]=1;if (!visit[i]) flag=0;if (line[i]%2!=0) count++;} if (flag) {if (count==2| | count==0) {printf ("yes\n");} else{printf ("no\n");}} Else{printf ("no\n");}}} return 0;}

nyoj-42-a stroke problem-dfs