1126. Eulerian Path (25)
2018-03-04 15:20
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1126. Eulerian Path (25)
时间限制300 ms内存限制65536 kB
代码长度限制16000 B
判题程序Standard作者CHEN, Yue
In graph theory, an Eulerian path is a path in a graph which visits every edge exactly once. Similarly, an Eulerian circuit is an Eulerian path which starts and ends on the same vertex. They were first discussed by Leonhard Euler while solving the famous Seven Bridges of Konigsberg problem in 1736. It has been proven that connected graphs with all vertices of even degree have an Eulerian circuit, and such graphs are called Eulerian. If there are exactly two vertices of odd degree, all Eulerian paths start at one of them and end at the other. A graph that has an Eulerian path but not an Eulerian circuit is called semi-Eulerian. (Cited from https://en.wikipedia.org/wiki/Eulerian_path)Given an undirected graph, you are supposed to tell if it is Eulerian, semi-Eulerian, or non-Eulerian.Input Specification:Each input file contains one test case. Each case starts with a line containing 2 numbers N (<= 500), and M, which are the total number of vertices, and the number of edges, respectively. Then M lines follow, each describes an edge by giving the two ends of the edge (the vertices are numbered from 1 to N).Output Specification:For each test case, first print in a line the degrees of the vertices in ascending order of their indices. Then in the next line print your conclusion about the graph -- either "Eulerian", "Semi-Eulerian", or "Non-Eulerian". Note that all the numbers in the first line must be separated by exactly 1 space, and there must be no extra space at the beginning or the end of the line.Sample Input 1:
7 12 5 7 1 2 1 3 2 3 2 4 3 4 5 2 7 6 6 3 4 5 6 4 5 6Sample Output 1:
2 4 4 4 4 4 2 EulerianSample Input 2:
6 10 1 2 1 3 2 3 2 4 3 4 5 2 6 3 4 5 6 4 5 6Sample Output 2:
2 4 4 4 3 3 Semi-EulerianSample Input 3:
5 8 1 2 2 5 5 4 4 1 1 3 3 2 3 4 5 3Sample Output 3:
3 3 4 3 3 Non-Eulerian注意判断图的连通性
为什么方法二测试点3 4 错误呢?希望dalao看出来能交流下...
明白了。。。 方法二图不连通忘记输出每个结点度了。。。
方法三对方法二修改。。。
方法一:#include<stdio.h>
#include<vector>
using namespace std;
int n,m,cnt;
vector<int>adj[510];
int degree[510];
int visit[510];
int cou=0;
void dfs(int s){
cnt++;
visit[s]=1;
int i;
for(i=0;i<adj[s].size();i++){
int next=adj[s][i];
if(visit[next]==0){
dfs(next);
}
}
}
void dfstravel(){
int i;
for(i=1;i<=n;i++){
if(visit[i]==0){
cou++;
dfs(i);
}
}
}
int main(){
int i,c1,c2;
scanf("%d %d",&n,&m);
for(i=0;i<m;i++){
scanf("%d %d",&c1,&c2);
adj[c1].push_back(c2);
adj[c2].push_back(c1);
degree[c1]++;
degree[c2]++;
}
for(i=1;i<=n;i++){
printf("%d",degree[i]);
if(i!=n){
printf(" ");
}
}
printf("\n");
int even=1,oddnum=0;
for(i=1;i<=n;i++){
if(degree[i]%2!=0){
even=0;
oddnum++;
}
}
dfstravel();
if(cou!=1){
printf("Non-Eulerian");return 0;
}
if(even){
printf("Eulerian");
}
else if(oddnum==2){
printf("Semi-Eulerian");
}
else{
printf("Non-Eulerian");
}
}方法二:测试点3 4错误#include<stdio.h>
int n,m,cou,cnt;
i
b337
nt degree[510];
int exist[510];
int visit[510];
int adj[510][510];
void dfs(int s){
if(visit[s]==1){
return;
}
visit[s]=1;
cou++;
for(int i=1;i<=n;i++){
if(adj[s][i]){
dfs(i);
}
}
}
int main(){
int i,c1,c2;
scanf("%d %d",&n,&m);
for(i=0;i<m;i++){
scanf("%d %d",&c1,&c2);
exist[c1]=exist[c2]=1;
adj[c1][c2]=adj[c2][c1]=1;
degree[c1]++;
degree[c2]++;
}
for(i=1;i<=n;i++){
if(exist[i]){//统计 出现过的点数
cnt++;
}
}
dfs(c1);
if(cou!=cnt){
printf("Non-Eulerian");
}
else{
int even=0,odd=0,oddindex;
for(i=1;i<=n;i++){
printf("%d",degree[i]);
if(i!=n){
printf(" ");
}
if(degree[i]%2){
odd++;
oddindex=i;
}
else{
even++;//
}
}
printf("\n");
if(odd==0){
printf("Eulerian");
}
else if(odd==2){//...
// cou=0;
// for(i=1;i<=n;i++){
// visit[i]=0;
// }
// dfs(oddindex);
// if(cou==n){
printf("Semi-Eulerian");
// }
// else{
// printf("Non-Eulerian");//测试点4
// }
}
else{
printf("Non-Eulerian");
}
}
}方法三:对方法二修改。。。#include<stdio.h>
int n,m,cou,cnt;
int degree[510];
int exist[510];
int visit[510];
int adj[510][510];
void dfs(int s){
if(visit[s]==1){
return;
}
visit[s]=1;
cou++;
for(int i=1;i<=n;i++){
if(adj[s][i]){
dfs(i);
}
}
}
int main(){
int i,c1,c2;
scanf("%d %d",&n,&m);
for(i=0;i<m;i++){
scanf("%d %d",&c1,&c2);
exist[c1]=exist[c2]=1;
adj[c1][c2]=adj[c2][c1]=1;
degree[c1]++;
degree[c2]++;
}
// for(i=1;i<=n;i++){
// if(exist[i]){//统计 出现过的点数
// cnt++;
// }
// }
dfs(1);
// if(cnt!=n);//while(1);//测试点4
// if(cou!=n){
// printf("Non-Eulerian");
// }
// else{
int even=1,odd=0;
for(i=1;i<=n;i++){
printf("%d",degree[i]);
if(i!=n){
printf(" ");
}
if(degree[i]%2){
odd++;
even=0;
}
}
printf("\n");
if(cou!=n){
printf("Non-Eulerian");return 0;
}
if(even){
printf("Eulerian");
}
else if(odd==2){
printf("Semi-Eulerian");
}
else{
printf("Non-Eulerian");
}
}
//}
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