程序設計大賽—文字遊戲

 

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給出一組單詞,判斷是否可以將單詞排序,使得每個單詞的第一個字母和前一個單詞的最後一個字母相同。

輸入格式:

每一行爲一整數 C ,表示測試數據的組數目。每組測試數據的第一行爲整數 N 1<=N<=100000 , 接下來的 N 行,每行爲一個單詞,每個單詞都只包含小寫字母並且最多包含 1000 個字符。

輸出格式:

如果不能將單詞序列重組以滿足要求,則輸出一行“ Impossible ”,否則輸出“ Possible .

輸入樣本:

3

2

ok

ok

2

ok

ko

3

Effects

Cse

Seu

輸出樣本:

Impossible

Possible

Possible

我的程序:

/***************************************************

程序名:文字遊戲

作者:許文發

時間: 2009-11-23

***************************************************/

#include<iostream.h>

#include<string.h>

#include<stdio.h>

int first=1;// 第一次輸出標誌, 1 代表是, 0 代表否

// 判斷整形數組 able 是否全爲 1 ,是則返回 1 ,否則返回 0

int isone(int able[],int num)

{

       int result=1;

       for(int i=0;i<num;i++)

       {

              if(able[i]==0)

                     result=0;

       }

       return result;

      

}

// 回溯算法

void words(char start[],char end[],int able[],int prev,int num)

{

       int j;

       if(isone(able,num) && able[num]==0)

       {

              able[num]=1;

       }

       else

       {

              for(j=0;j<num;j++)

              {

                     if(end[prev]==start[j] && able[j]==0)

                     {

                            able[j]=1;

                            words(start,end,able,j,num);

                            able[j]=0;

                     }

              }

       }

}

// 清空文件

void clearfile()

{

       FILE *pt;

       pt=fopen("output.txt","w");

       fclose(pt);

}

// 寫文件

void mywrite(int n,int first)

{

       FILE *pt;

       pt=fopen("output.txt","a");

       if(first==1)

       {

              if(n==1)

                     fprintf(pt,"Possible");

              else

                     fprintf(pt,"Impossible");

       }

       else

       {

              if(n==1)

                     fprintf(pt,"/nPossible");

              else

                     fprintf(pt,"/nImpossible");

       }

       fclose(pt);

}

// 主函數

void main()

{

       clearfile();

       FILE *pt;

       char aa[1000];

       char start[100000],end[100000];

       int able[100000];

       int i,ii;

       int total,m;

       if(NULL==(pt=fopen("input.txt","r")))

       {

       }

       else

       {

              fscanf(pt,"%d",&total);

              for(ii=0;ii<total;ii++)

              {

                     fscanf(pt,"%d",&m);

                     for(i=0;i<100000;i++)

                            able[i]=0;

                     for(i=0;i<m;i++)

                     {

                            fscanf(pt,"%s",aa);

                            start[i]=aa[0];

                            end[i]=aa[strlen(aa)-1];

                     }

                     for(i=0;i<m;i++)

                     {

                            if(able[m]==0)

                            {

                                   able[i]=1;

                                   words(start,end,able,i,m);

                                   able[i]=0;

                            }

                            else

                                   break;

                     }

            mywrite(able[m],first);

                     first=0;

              }

              fclose(pt);

       }

}

輸入:

3

2

ok

eo

13

ege

hhs

sign

nine

edit

two

one

enn

text

th

high

sws

sst

2

ok

ok

輸出:

Possible

Possible

Impossible

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