hdu 1116 并查集判断欧拉回路通路

判断一些字符串能首尾相连连在一起

并查集求欧拉回路和通路

Sample Input

3

2

acm

ibm

3

acm

malform

mouse

2

ok

ok

Sample Output

The door cannot be opened.

Ordering is possible.

The door cannot be opened.

 1 #include<cstdio>

 2 #include<iostream>

 3 #include<algorithm>

 4 #include<cstring>

 5 #include<cmath>

 6 #include<queue>

 7 #include<map>

 8 using namespace std;

 9 #define MOD 1000000007

10 const int INF=0x3f3f3f3f;

11 const double eps=1e-5;

12 typedef long long ll;

13 #define cl(a) memset(a,0,sizeof(a))

14 #define ts printf("*****\n");

15 const int MAXN=1005;

16 int n,m,tt;

17 char s[MAXN];

18 int in[27],out[27],vis[27],f[27],p[27];

19 int find(int x)

20 {

21     if(f[x]==-1)return x;

22     return f[x]=find(f[x]);

23 }

24 void bing(int x,int y)

25 {

26     int t1=find(x);

27     int t2=find(y);

28     if(t1!=t2)

29     {

30         f[t1]=t2;

31     }

32 }

33 void init()

34 {

35     cl(vis);

36     cl(in);

37     cl(out);

38     memset(f,-1,sizeof(f));

39 }

40 int main()

41 {

42     int i,j,k;

43     #ifndef ONLINE_JUDGE

44     freopen("1.in","r",stdin);

45     #endif

46     scanf("%d",&tt);

47     while(tt--)

48     {

49         init();

50         scanf("%d",&n);

51         for(i=0;i<n;i++)

52         {

53             scanf("%s",s);

54             int len=strlen(s);

55             int x=s[0]-'a';

56             int y=s[len-1]-'a';

57             out[x]++;

58             in[y]++;

59             bing(x,y);

60             vis[x]=vis[y]=1;

61         }

62         int cnt=0;  //统计个数

63         for(i=0;i<26;i++)

64         {

65             if(vis[i]&&find(i)==i)

66             {

67                 cnt++;

68             }

69         }

70         if(cnt>1)

71         {

72             printf("The door cannot be opened.\n");

73             continue;

74         }

75         int tot=0;  //统计出入度不相同的点

76         for(i=0;i<26;i++)

77         {

78             if(vis[i]&&in[i]!=out[i])

79             {

80                 p[tot++]=i;

81             }

82         }

83         if(tot==0)

84         {

85             printf("Ordering is possible.\n");

86             continue;

87         }

88         if(tot==2&&((out[p[0]]-in[p[0]]==1&&in[p[1]]-out[p[1]]==1)||(out[p[1]]-in[p[1]]==1&&in[p[0]]-out[p[0]]==1))) //欧拉通路

89         {

90             printf("Ordering is possible.\n");

91             continue;

92         }

93         printf("The door cannot be opened.\n");

94     }

95 }