HDU 1878 Euro-pull circuit "and check set + Eulerian graph"

Euro-Pull circuit Time limit:2000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)
Total submission (s): 9655 Accepted Submission (s): 3498


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 graph, ask if there is a Euler loop.
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 3 1 2 1 3 2 3 3 2 1 2 2 3 0
Sample Output

1 0/*
The problem of Euler loops based on the Eulerian graph graph is that the condition of a graph is: 0 of all nodes have an odd number of degrees
1. Use and check the set to determine whether a graph
2. Determine if all the nodes are in an even number of degrees
*/

#include <cstdio> #include <cstring> #include <vector> using namespace std; 

int pre[1005],in[1005];
    struct Edge {int from;
int to;
}E[100005];

int find (int x) {return x==pre[x]?x:pre[x]=find (pre[x]);}
    void join (int x,int y) {int fx=find (x), Fy=find (y);
    if (fx!=fy) {pre[fx]=fy;
    }} int main () {int n,m,i,t,flag;
        while (scanf ("%d", &n) &&n) {scanf ("%d", &m);
        int ok=0;
        memset (In,0,sizeof (in));
        memset (E,0,sizeof (e));
        for (I=1; i<=n; i++) pre[i]=i;
            for (I=1; i<=m; i++) {scanf ("%d%d", &e[i].from,&e[i].to);
        in[e[i].from]++,in[e[i].to]++;//statistics into join (E[I].FROM,E[I].TO);
            } for (i=1,t=flag=0; i<=n; i++) {if (in[i]&1) t++;  
        if (pre[i]==i) flag++;
        } if (flag==1&&t==0) ok=1;
        if (OK) printf ("1\n");
    else printf ("0\n"); } return 0; }