题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1116
思路:先合并,然后判断根结点是否唯一,如果是,则在判断是否是欧拉路;否则,直接结束。
View Code
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 using namespace std; 5 const int MAXN=100000+10; 6 int In[MAXN],Out[MAXN]; 7 int parent[MAXN]; 8 bool mark[MAXN]; 9 int n; 10 11 12 void Initiate(){ 13 for(int i=0;i<n;i++){ 14 parent[i]=-1; 15 } 16 } 17 18 int Find(int x){ 19 int s; 20 for(s=x;parent[s]>=0;s=parent[s]); 21 while(s!=x){ 22 int tmp=parent[x]; 23 parent[x]=s; 24 x=tmp; 25 } 26 return s; 27 } 28 29 void Union(int R1,int R2){ 30 int r1=Find(R1); 31 int r2=Find(R2); 32 if(r1!=r2){ 33 parent[r1]=r2; 34 } 35 } 36 37 int main(){ 38 int _case; 39 scanf("%d",&_case); 40 while(_case--){ 41 scanf("%d",&n); 42 Initiate(); 43 memset(mark,false,sizeof(mark)); 44 memset(In,0,sizeof(In)); 45 memset(Out,0,sizeof(Out)); 46 for(int i=0;i<n;i++){ 47 char str[1010]; 48 scanf("%s",str); 49 int x=str[0]-'a'; 50 int y=str[strlen(str)-1]-'a'; 51 Union(x,y); 52 mark[x]=mark[y]=true; 53 Out[x]++; 54 In[y]++; 55 } 56 int count=0; 57 for(int i=0;i<26;i++){ 58 if(mark[i]&&parent[i]==-1){ 59 count++; 60 } 61 } 62 if(count>1){ 63 printf("The door cannot be opened.\n"); 64 }else { 65 int p[27]; 66 int k=0; 67 for(int i=0;i<26;i++){ 68 if(mark[i]&&In[i]!=Out[i]){ 69 p[k++]=i; 70 } 71 } 72 if(k==0){ 73 printf("Ordering is possible.\n"); 74 }else { 75 if(k==2&&abs(In[p[0]]-Out[p[0]])==1&&abs(In[p[1]]-Out[p[1]])==1){ 76 printf("Ordering is possible.\n"); 77 }else 78 printf("The door cannot be opened.\n"); 79 } 80 } 81 } 82 return 0; 83 }