Hdoj Euler circuit 1878 "Euler Loop Determination"

Euler circuit Time limit:2000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)
Total submission (s): 10459 Accepted Submission (s): 3815


Problem description is a loop that does not leave the pen on the paper, but only once per side of the picture, and can return to the starting point. Now given a diagram, ask if there is a Euler circuit?
The input test inputs contain several test cases. The 1th line of each test case gives two positive integers, namely the number of nodes N (1 < N < 1000) and the number of sides m, followed by M-bars, each with a pair of positive integers, the number of two nodes (nodes from 1 to N) that are directly connected to the bar. When n is 0 o'clock input knot
Beam.
Output takes one row for each test case, and outputs 1 if the Euler loop is present, otherwise outputs 0.

Sample Input

3 31 21 32 33 21 22 30

Sample Output

10

Authorzju
SOURCE Zhejiang University Computer Postgraduate exam on the machine-2008
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#include <iostream> #include <stdio.h> #include <string.h> #include <algorithm> #include < Vector>using namespace Std;int d[1010];int r[1010];int find_x (int x) {int son=x;int temp;while (x!=r[x]) x=r[x];while ( son!=x) {temp=r[son];r[temp]=x;son=temp;} return x;} int fun (int x,int y) {x=find_x (x); y=find_x (y); if (x!=y) {r[x]=y;}} int main () {int n,m;while (scanf ("%d", &n), N) {scanf ("%d", &m), memset (d,0,sizeof (d)); int a,b;for (int i=0;i<= n;i++) {r[i]=i;} for (int i=0;i<m;i++) {scanf ("%d%d", &a,&b), fun (A, b);d [a]++;d [b]++;} int flag=1;int num=0;for (int i=1;i<=n;i++) {if (r[i]==i) num++;} for (int i=1;i<=n;i++) {if (d[i]&1) {flag=0;break;}} if (flag&&num==1) printf ("1\n"), Else printf ("0\n");}    return 0;}

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Hdoj Euler circuit 1878 "Euler Loop Determination"