POJ1300 Door Man Euler Loop judgment, poj1300door

POJ1300 Door Man Euler Loop judgment, poj1300door

Question link:

1300

Question:

In a house (No. 0 ~ N-1) N rooms and X doors connecting two rooms, all rooms are connected, and the door will be closed every time a door passes. Q: Can a person depart from room M successfully reach room 0 and close all doors.

Question:

This is an entry to the Euler loop. First, we will learn the judgment theorem of the undirected Euler Loop:

An undirected graph G has an Euler's path. The condition is that G is a connected graph, and G has only two odd-degree nodes (vertices with odd numbers) or no odd-degree nodes.

This is an undirected graph with a fixed start point and end point.

Therefore, there are only two situations that can constitute the Euler Loop:

1 The degrees of all vertices are even

2. the start point and end points are odd and the other values are even.

Code:

#include<iostream>#include<cstdio>#include<cstring>using namespace std;int main(){    int degree[30];    int m,n,i,j,s,odd,even;    char str[1005],str2[1005];    while(scanf("%s",str)&&strcmp(str,"ENDOFINPUT")!=0)    {        s=odd=even=0;        memset(degree,0,sizeof(degree));        scanf("%d%d",&m,&n);        getchar();        for(i=0; i<n; i++)        {            gets(str2);            for(j=0; j<strlen(str2); j++)                if(str2[j]>='0'&&str2[j]<='9')                {                    s++;                    degree[i]++;                    degree[str2[j]-'0']++;                }        }        scanf("%s",str);        for(i=0; i<n; i++)        {            if(degree[i]%2==0)                even++;            else                odd++;        }        if((odd==0&&m==0)||(odd==2&&degree[m]%2==1&&degree[0]%2==1&&m!=0))            cout<<"YES"<<' '<<s<<endl;        else            cout<<"NO"<<endl;    }    return 0;}