POJ–1300–Door Man【推断无向图欧拉通路】

链接：http://poj.org/problem?id=1300

题意：有n个房间。每一个房间有若干个门和别的房间相连。管家从m房间開始走。要回到自己的住处（0），问是否有一条路能够走遍全部的门而且没有反复的路。

无向图欧拉通路充要条件：G为连通图，而且G仅有两个奇度结点（度数为奇数的顶点）或者无奇度结点。

无向图欧拉回路充要条件：G为无奇度结点的连通图。

思路：推断是否存在欧拉通路。依据欧拉通路、欧拉回路的性质来做。有两种情况：一种是欧拉回路。全部房间的门的个数都是偶数个，而且此时初始房间不是0，此时存在要求的路径。假设初始是0则不行。还有一种是欧拉通路。仅仅有两个房间门是奇数个。剩下都是偶数个。而且这两个房间一个是0。一个是当前起点，而且起点不能是0，此时也存在要求的路径，否则不存在。

输入比較蛋疼

#include<cstring>
#include<string>
#include<fstream>
#include<iostream>
#include<iomanip>
#include<cstdio>
#include<cctype>
#include<algorithm>
#include<queue>
#include<map>
#include<set>
#include<vector>
#include<stack>
#include<ctime>
#include<cstdlib>
#include<functional>
#include<cmath>
using namespace std;
#define PI acos(-1.0)
#define MAXN 500100
#define eps 1e-7
#define INF 0x7FFFFFFF
#define LLINF 0x7FFFFFFFFFFFFFFF
#define seed 131
#define mod 1000000007
#define ll long long
#define ull unsigned ll
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1

int in[30],out[30];
char s[20],s1[200];
int main(){
    int i,j;
    int m,n;
    int a,flag,ans,fk;
    while(scanf("%s",s)!=EOF){
        if(s[0]=='E'&&strlen(s)>5)  break;
        scanf("%d%d",&m,&n);
        getchar();
        ans = 0;
        flag = 0;
        fk = 0;
        memset(in,0,sizeof(in));
        for(i=0;i<n+1;i++){
            gets(s1);
            int p = 0;
            while(sscanf(s1+p,"%d",&a)==1){
                ans++;
                in[a]++;
                in[i]++;
                while(s1[p]!='
#include<cstring> #include<string> #include<fstream> #include<iostream> #include<iomanip> #include<cstdio> #include<cctype> #include<algorithm> #include<queue> #include<map> #include<set> #include<vector> #include<stack> #include<ctime> #include<cstdlib> #include<functional> #include<cmath> using namespace std; #define PI acos(-1.0) #define MAXN 500100 #define eps 1e-7 #define INF 0x7FFFFFFF #define LLINF 0x7FFFFFFFFFFFFFFF #define seed 131 #define mod 1000000007 #define ll long long #define ull unsigned ll #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 int in[30],out[30]; char s[20],s1[200]; int main(){ int i,j; int m,n; int a,flag,ans,fk; while(scanf("%s",s)!=EOF){ if(s[0]=='E'&&strlen(s)>5) break; scanf("%d%d",&m,&n); getchar(); ans = 0; flag = 0; fk = 0; memset(in,0,sizeof(in)); for(i=0;i<n+1;i++){ gets(s1); int p = 0; while(sscanf(s1+p,"%d",&a)==1){ ans++; in[a]++; in[i]++; while(s1[p]!='\0'&&s1[p]!=' ') p++; while(s1[p]!='\0'&&s1[p]==' ') p++; } } for(i=0;i<n;i++){ if(in[i]&1) flag++; } if(!flag){ if(!m) fk = 1; else fk = 0; } else{ if(flag==2&&m!=0&&in[m]&1&&in[0]&1) fk = 1; else fk = 0; } if(fk) printf("YES %d\n",ans); else puts("NO"); } return 0; }
'&&s1[p]!=' ')  p++;
                while(s1[p]!='
#include<cstring> #include<string> #include<fstream> #include<iostream> #include<iomanip> #include<cstdio> #include<cctype> #include<algorithm> #include<queue> #include<map> #include<set> #include<vector> #include<stack> #include<ctime> #include<cstdlib> #include<functional> #include<cmath> using namespace std; #define PI acos(-1.0) #define MAXN 500100 #define eps 1e-7 #define INF 0x7FFFFFFF #define LLINF 0x7FFFFFFFFFFFFFFF #define seed 131 #define mod 1000000007 #define ll long long #define ull unsigned ll #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 int in[30],out[30]; char s[20],s1[200]; int main(){ int i,j; int m,n; int a,flag,ans,fk; while(scanf("%s",s)!=EOF){ if(s[0]=='E'&&strlen(s)>5) break; scanf("%d%d",&m,&n); getchar(); ans = 0; flag = 0; fk = 0; memset(in,0,sizeof(in)); for(i=0;i<n+1;i++){ gets(s1); int p = 0; while(sscanf(s1+p,"%d",&a)==1){ ans++; in[a]++; in[i]++; while(s1[p]!='\0'&&s1[p]!=' ') p++; while(s1[p]!='\0'&&s1[p]==' ') p++; } } for(i=0;i<n;i++){ if(in[i]&1) flag++; } if(!flag){ if(!m) fk = 1; else fk = 0; } else{ if(flag==2&&m!=0&&in[m]&1&&in[0]&1) fk = 1; else fk = 0; } if(fk) printf("YES %d\n",ans); else puts("NO"); } return 0; }
'&&s1[p]==' ')  p++;
            }
        }
        for(i=0;i<n;i++){
            if(in[i]&1) flag++;
        }
        if(!flag){
            if(!m)  fk = 1;
            else    fk = 0;
        }
        else{
            if(flag==2&&m!=0&&in[m]&1&&in[0]&1)   fk = 1;
            else    fk = 0;
        }
        if(fk)  printf("YES %d\n",ans);
        else    puts("NO");
    }
    return 0;
}