poj1386 字符串类型的一笔画问题 欧拉回路

Play on Words

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 10685		Accepted: 3625

Description

Some of the secret doors contain a very interesting word puzzle. The team of archaeologists has to solve it to open that doors. Because there is no other way to open the doors, the puzzle is very important for us.

There is a large number of magnetic plates on every door. Every
plate has one word written on it. The plates must be arranged into a
sequence in such a way that every word begins with the same letter as
the previous word ends. For example, the word ``acm'' can be followed by
the word ``motorola''. Your task is to write a computer program that
will read the list of words and determine whether it is possible to
arrange all of the plates in a sequence (according to the given rule)
and consequently to open the door.

Input

The
input consists of T test cases. The number of them (T) is given on the
first line of the input file. Each test case begins with a line
containing a single integer number Nthat indicates the number of plates
(1 <= N <= 100000). Then exactly Nlines follow, each containing a
single word. Each word contains at least two and at most 1000 lowercase
characters, that means only letters 'a' through 'z' will appear in the
word. The same word may appear several times in the list.

Output

Your
program has to determine whether it is possible to arrange all the
plates in a sequence such that the first letter of each word is equal to
the last letter of the previous word. All the plates from the list must
be used, each exactly once. The words mentioned several times must be
used that number of times.

If there exists such an ordering of plates, your program should
print the sentence "Ordering is possible.". Otherwise, output the
sentence "The door cannot be opened.".

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.

Source

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
int edge[][];
struct node{
   int in,out;
}que[];
int  vis[];
char str[];
int cnt_dfs;

void dfs(int u){
    vis[u]=;
    for(int i=;i<;i++)
        if(!vis[i]&&edge[u][i])
            dfs(i);

            cnt_dfs++;
}

int main(){
   int T;
   scanf("%d",&T);

   while(T--){
    int n;
    scanf("%d",&n);
    memset(edge,,sizeof(edge));
    memset(vis,,sizeof(vis));
    memset(que,,sizeof(que));
    int u,v;
   // getchar();

      for(int i=;i<=n;i++){

        scanf("%s",str);
      //  getchar();
        int len=strlen(str);
         u=str[]-'a';
         v=str[len-]-'a';
         que[u].out++;
         que[v].in++;
         edge[u][v]=edge[v][u]=;

      }

      int cnt=,temp=;
   //  bool flag=false;
       for(int i=;i<;i++){
        if(que[i].in||que[i].out){
            if(!temp){

                temp=i;
            }
            cnt++;
        }
       }
  //  printf("--->%d\n",temp);
      cnt_dfs=;
      dfs(temp);
    // printf("-->%d %d\n",cnt,cnt_dfs);

      if(cnt_dfs!=cnt){
        printf("The door cannot be opened.\n");
        continue;
      }
     // bool flag=false;
      int cnt_s=,cnt_e=,cnt_m=;
      cnt=;
      for(int i=;i<;i++){
        if(que[i].in==&&que[i].out==)
            continue;
            cnt++;
        if(que[i].in==que[i].out)
           cnt_m++;
        else if(que[i].in-que[i].out==-)
            cnt_s++;
        else  if(que[i].in-que[i].out==)
            cnt_e++;
      }

      if(cnt_m==cnt||(cnt_m==cnt-&&cnt_s==&&cnt_e==))
        printf("Ordering is possible.\n");
      else
        printf("The door cannot be opened.\n");

   }
   return ;
}