SCU 4520 Euler-Euler circuit

EulerTime limit:0MS Memory Limit:0KB 64bit IO Format:%lld &%llu

Description

Time limit:1000 MS Memory limit:256 M

A graph of n points and m edges is given, and it is judged whether there is a Oraton path under the condition of both the graph and the direction graph.

Input

The input contains multiple sets of data. The first behavior is an integer t (1?≤? T?≤?100), representing the number of data groups, for each group of data: The first row is two integers n and m ( 1?≤? n? ≤?500,?0?≤? m≤? N (n? −?1)/2), respectively, representing the number of points on the graph and the number of edges.
Then the M line, two integers per line, ui and vi ( 1?≤? Ui,? v i? ≤? n,? u i? ≠? v i ), which represents the two points connected by an edge on the diagram. The input guarantees no heavy edges.

Output

First Judge: If this picture is a non-direction diagram, whether there is Euler pathway;
Second judgment: If this picture is a graph, whether there is Oraton road.
For each judgment, if present, output "Yes", otherwise output "no" (not including quotation marks). Two judgments are separated by a space.

Sample input

3

2 1
1 2

4 3
1 2
1 3
1 4

4 4
1 2
1 3
1 4
2 3

Sample output

Yes Yes
No no
Yes No

Hint

Euler's access, Eulerian graph circuit,
no-show diagram :
1) Set G is a connected undirected graph, it is said that through the g of each edge once and only once the path is Euler path;
2) If the Oraton road is a loop (the starting and ending points are the same vertex), then this loop is called the Euler circuit (circuit);
3) The non-direction graph G with a Euler loop is called Eulerian graph (Euler graph).
to map :
1) Set D is the direction graph, the base diagram of D is connected, it is said that each edge of D is once and only once of the direction of the direction of the Euler path;
2) If there is a forward loop to the Oraton road, it is said that there is a forward loop to the Euler circuit (directed Euler circuit);
3) A direction diagram D with a forward Euler loop is called a Eulerian graph (directed Euler graph).

Extend

Euler loop print path algorithm: Fleury (Flore) algorithm

Author

Goozy

It can be done in degrees.

#include <iostream> #include <cstring> #include <cstdio> #include <algorithm> #include < queue> #include <vector> #include <iomanip> #include <math.h> #include <map>using namespace STD; #define FIN freopen ("Input.txt", "R", stdin), #define FOUT freopen ("Output.txt", "w", stdout); #define INF 0x3f3f 3f3f#define Lson l,m,rt<<1#define Rson m+1,r,rt<<1|1typedef long long ll;const int MAXN=1000+5;int DU[MA   Xn];int in[maxn],out[maxn];int n,m,sz;int father[maxn];void edge_init () {memset (du,0,sizeof (DU));   memset (In,0,sizeof (in));   Memset (out,0,sizeof (out));   for (int i=1;i<=n;i++) {father[i]=i;    }}int Find (int x) {if (x!=father[x]) {father[x]=find (father[x]); } return father[x];}    int main () {//fin int T;    scanf ("%d", &t);        while (t--) {scanf ("%d%d", &n,&m);        Edge_init ();        int sum=n;            for (int i=1;i<=m;i++) {int u,v; scanf ("%d%d", &AMP;U,&AMP;V);            du[u]++;            du[v]++;            in[v]++;            out[u]++;            int root1=find (u);            int Root2=find (v);                if (Root1!=root2) {father[root2]=root1;            sum--;            }} if (Sum!=1) {printf ("No no\n");        Continue        } int cnt=0;        for (int i=1;i<=n;i++) {if (du[i]%2==1) cnt++; } if (cnt==0| |        cnt==2) printf ("Yes");        else printf ("No");            BOOL Flag=1;            int fir=0, sec=0;                    for (int i=1;i<=n;i++) {if (In[i]!=out[i]) {if (!fir&&in[i]-out[i]==1) fir=1;                    else if (!sec&&in[i]-out[i]==-1) {sec=1;                    }else{flag=0;            }}} if (flag) printf ("yes\n");    else printf ("no\n"); } return 0;}

　　

SCU 4520 Euler-Euler circuit