Euler circuit base HDU1878 Euler circuit | | and the difference set

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Euler circuit

Time limit:2000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)
Total submission (s): 10544 Accepted Submission (s): 3845

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 INPUT3 31 21 32 33 21 22 30

Sample Output10

Authorzju

Source Zhejiang University Computer Postgraduate exam on the machine-2008

Recommendwe has carefully selected several similar problems for you:1877 1881 1863 1875 1864 attached: PS naked Euler circuit, below two No The gaze was written by me, without gaze, as a template, with a gaze on the web, easy to understand thanks to the great God

#include <stdio.h>#include<string.h>#include<iostream>#include<algorithm>using namespacestd;Const intmaxn=1005;intMAP[MAXN][MAXN];BOOLVIS[MAXN];intINDEGREE[MAXN];intn,m;voidDfsintu) {Vis[u]=true;  for(intI=1; i<=n;i++)    if(!vis[i]&&Map[u][i]) DFS (i); return ;}intMain () { while(SCANF ("%d", &n) &&N) {scanf ("%d",&m); memset (Map,0,sizeof(map)); memset (Vis,false,sizeof(VIS)); memset (Indegree,0,sizeof(Indegree));  for(intI=0; i<m;i++){            intu,v; scanf ("%d%d",&u,&v); Indegree[u]++; INDEGREE[V]++; MAP[U][V]=map[v][u]=1; } DFS (1); inti;  for(i=1; i<=n;i++){            if(!vis[i]| | indegree[i]%2==1)                 Break; }        if(i>N) printf ("1\n"); Elseprintf ("0\n"); }    return 0;}

//This paper mainly examines the two necessary and sufficient conditions of the Euler circuit: 1. Unicom Figure 2. Vertex degrees are even (two conditions are indispensable)#include<stdio.h>#include<string.h>intgraph[ +][ +];//using adjacency matrix to store graphsintvisit[ +];//Mark whether the point has been accessed over the durationintdegree[ +];//the degree of storage nodevoidDFS (intVintN) {//Depth-first traversal, recursiveVISIT[V] =1;  for(intI=1; i<=n; i++)        if(Graph[v][i] && visit[i]==0) DFS (i,n);}intMain () {//freopen ("In.txt", "R", stdin);    intN, M;  while(SCANF ("%d%d", &n,&m)!=eof &&N) {memset (graph,0,sizeof(graph));//Qing 0memset (Visit,0,sizeof(visit)); memset (degree,0,sizeof(degree)); intA, B, I; intFlag =1;//Mark if there is a Euler loop         for(i=0; i<m; i++) {scanf ("%d%d",&a,&b); GRAPH[A][B]= Graph[b][a] =1;//"1" indicates that two points belong to the adjacency relationshipdegree[a]++; degree[b]++; } DFS (1, N);  for(i=1; i<=n; i++) {            if(Visit[i] = =0) {flag=0;//If a bit has not been accessed, that is, a deep traversal with an unreachable point, there is no Euler loop                 Break; }            if(Degree[i]%2!=0) {flag=0;//if the degree is not even, then the Euler circuit does not exist.                 Break; }        }        if(flag) printf ("1\n"); Elseprintf ("0\n"); }    return 0;}

Euler circuit base HDU1878 Euler circuit | | and the difference set