NYoj-42-one stroke problem-DFS

NYoj-42-one stroke problem-DFS
One stroke problem time limit: 3000 MS | memory limit: 65535 KB difficulty: 4

Description

Zyc prefers to play games from an early age, including drawing a stroke. He wants to ask you to write a program for him to determine whether a graph can be used.

It is specified that all edges can be painted only once and cannot be repeated.

Input

The first row only has a positive integer N (N <= 10) indicating the number of test data groups.
The first row of each group of test data has two positive integers P, Q (P <= 1000, Q <= 2000), indicating the number of vertices and connections in the painting respectively. (The vertex number ranges from 1 to P)
In the subsequent Q rows, each row has two positive integers A and B (0 output

If there is a line that meets the conditions, the output is "Yes ",
If No matching line exists, output "No ".

Sample Input

24 31 21 31 44 51 22 31 31 43 4

Sample output

NoYes

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              Using namespace std; int line [1010]; // count the degree of each node int visit [1010]; // records the accessed point int G [1010] [1010]; // record I, j whether to connect int vertex, edge; void DFS (int I) // determine whether to connect {// int v = I; visit [I] = 1; // I accessed for (int v = 0; v <= vertex; v ++) {if (v! = I & G [I] [v] &! Visit [v]) // find one! = I/connect to I/No accessed point DFS (v); // continue to access from this point} int main () {int T, x, y, count, flag; while (~ Scanf ("% d", & T) {while (T --) {count = 0; flag = 1; scanf ("% d", & vertex, & edge); // Number of vertices and edges memset (line, 0, sizeof (line); memset (visit, 0, sizeof (visit); memset (G, 0, sizeof (G); for (int I = 1; I <= edge; I ++) {scanf ("% 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 ;}