hdu 1325 判断有向图是否为树

题意：判断有向图是否为树

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这题用并查集判断连通，连通后有且仅有1个入度为0，其余入度为1，就是树了

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

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

14 const int MAXN=1000000;

15 int n,m,tt;

16 int a[MAXN];

17 int f[MAXN];

18 int in[MAXN];

19 bool vis[MAXN];

20 int find(int x)

21 {

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

23     else return f[x]=find(f[x]);

24 }

25 void bing(int x,int y)

26 {

27     int t1=find(x);

28     int t2=find(y);

29     if(t1!=t2)

30         f[t1]=t2;

31 }

32 int main()

33 {

34     int i,j,k;

35     #ifndef ONLINE_JUDGE

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

37     #endif

38     int cnt=0;

39     int u,v;

40     int iCase=0;

41     bool flag=true;

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

43     memset(vis,false,sizeof(vis));

44     memset(in,0,sizeof(in));

45     while(scanf("%d%d",&u,&v)!=EOF)

46     {

47         if(u==-1&&v==-1)    break;

48         if(u==0&&v==0)

49         {

50             iCase++;

51             printf("Case %d ",iCase);

52             int t0=0;

53             for(i=0;i<cnt;i++)

54             {

55                 if(find(a[i])!=find(a[0]))

56                 {

57                     flag=0;

58                     break;

59                 }

60                 if(in[a[i]]==0) t0++;

61                 else if(in[a[i]]>1)

62                 {

63                     flag=0;

64                     break;

65                 }

66             }

67             if(t0!=1)flag=false;

68             if(cnt==0)flag=true;

69             if(flag)printf("is a tree.\n");

70             else printf("is not a tree.\n");

71             cnt=0;

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

73             memset(vis,false,sizeof(vis));

74             memset(in,0,sizeof(in));

75             flag=true;

76         }

77         else

78         {

79             if(!vis[u])

80             {

81                 vis[u]=1;

82                 a[cnt++]=u;

83             }

84             if(!vis[v])

85             {

86                 vis[v]=1;

87                 a[cnt++]=v;

88             }

89             if(find(u)==find(v))    flag=0;

90             else

91             {

92                 bing(u,v);

93                 in[v]++;

94             }

95         }

96     }

97 }