poj 1386

Play on Words

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 7877 Accepted: 2778

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

32acmibm3acmmalformmouse2okok

Sample Output

The door cannot be opened.Ordering is possible.The door cannot be opened.

题意：类似成语接龙，前一个的最后一个字母与下一个的开始要相同，问能否将所给的单词一个接一个全部连起来？

思路：欧拉回路的判定。

#include<iostream>#include<stdio.h>#include<string.h>using namespace std;#define M 27int in[M],out[M];//入度，出度int p[M];//并查集数组bool used[M];char str[1010];void init(){    for(int i=0; i<26; i++)    {        in[i]=0;        out[i]=0;        used[i]=false;        p[i]=-1;    }}int find(int x){    if(p[x]>=0)    {        p[x]=find(p[x]);        return p[x];    }    return x;}void unio(int a,int b){    int x=find(a);    int y=find(b);    if(x==y)        return ;    int xx=p[x];    int yy=p[y];    if(xx<yy)    {        p[y]=x;        p[x]+=yy;    }    else    {        p[x]=y;        p[y]+=xx;    }}int main(){    int cas=0,st=0,end=0,len=0,i=0;    int n=0;    scanf("%d",&cas);    while(cas--)    {        init();        scanf("%d",&n);        for(i=0; i<n; i++)        {            scanf("%s",str);            len=strlen(str);            st=str[0]-'a';            end=str[len-1]-'a';            in[end]++;            out[st]++;            used[st]=1;            used[end]=1;            unio(st,end);        }        int scc=0;        for(i=0; i<26; i++)        {            if(used[i]&&p[i]<0)                scc++;        }        if(scc>1)        {            printf("The door cannot be opened.\n");            continue;        }        int pp=0,q=0;        for(i=0; i<26; i++)        {            if(used[i]&&in[i]!=out[i])            {                if(1==(in[i]-out[i]))                    pp++;                else if(1==(out[i]-in[i]))                    q++;                else                    break;            }        }        if(i<25)//说明中间有点的出入度数差值不是1或者是0            printf("The door cannot be opened.\n");        else if((0==pp&&0==q)||(1==pp&&1==q))            printf("Ordering is possible.\n");        else            printf("The door cannot be opened.\n");    }    return 0;}/*32acmibm3acmmalformmouse2okok*/