Sorting It All Out 拓扑排序&&判断是否存在环，是否关系包含所有的点

Problem Description

An ascending sorted sequence of distinct values is one in which some form of a less-than operator is used to order the elements from smallest to largest. For example, the sorted sequence A, B, C, D implies that A < B, B < C and C < D. in this problem, we will give you a set of relations of the form A < B and ask you to determine whether a sorted order has been specified or not.

 

Input

Input consists of multiple problem instances. Each instance starts with a line containing two positive integers n and m. the first value indicated the number of objects to sort, where 2 <= n <= 26. The objects to be sorted will be the first n characters of the uppercase alphabet. The second value m indicates the number of relations of the form A < B which will be given in this problem instance. Next will be m lines, each containing one such relation consisting of three characters: an uppercase letter, the character "<" and a second uppercase letter. No letter will be outside the range of the first n letters of the alphabet. Values of n = m = 0 indicate end of input.

 

Output

For each problem instance, output consists of one line. This line should be one of the following three: 

Sorted sequence determined after xxx relations: yyy...y. 
Sorted sequence cannot be determined. 
Inconsistency found after xxx relations. 

where xxx is the number of relations processed at the time either a sorted sequence is determined or an inconsistency is found, whichever comes first, and yyy...y is the sorted, ascending sequence. 

 

Sample Input

4 6

A<B

A<C

B<C

C<D

B<D

A<B

3 2

A<B

B<A

26 1

A<Z

0 0

 

Sample Output

Sorted sequence determined after 4 relations: ABCD.

Inconsistency found after 2 relations.

Sorted sequence cannot be determined.

**************************************************************************************************************************经典的拓扑排序

***************************************************************************************************************************

  1 #include<iostream>
  2 #include<string>
  3 #include<cstring>
  4 #include<cstdio>
  5 #include<queue>
  6 using namespace std;
  7 int a[26][26];
  8 queue<char>Q1;//记录两者之间的关系
  9 int de[26],de2[26];
 10 int relations;
 11 bool flag;
 12 int kind,it,jt,kt,n,m;
 13 void carry(int n,int ret)
 14 {
 15   int i,j,sum=0;
 16   int num;
 17   bool sign=true;
 18   queue<int>Q;
 19   for(i=0;i<n;i++)
 20   {
 21       if(de[i]==0)
 22         Q.push(i);
 23   }
 24   while(!Q.empty())
 25   {
 26       num=0;
 27      for(i=0;i<n;i++)
 28      {
 29          if(de[i]==0)
 30            num++;
 31      }
 32      if(num>1)
 33        sign=false;//存在不确定关系
 34      i=Q.front();
 35      Q.pop();
 36      Q1.push(i+'A');//
 37      sum++;
 38      de[i]=-1;
 39      for(j=0;j<n;j++)
 40      {
 41          if(a[i][j]==true)
 42          {
 43              de[j]--;
 44              if(de[j]==0)
 45               Q.push(j);
 46          }
 47      }
 48 
 49 
 50   }
 51   //如果存在确定关系，记录位置，不再往下算
 52   if(sum==n&&sign)
 53   {
 54       flag=true;
 55       kind=1;
 56       relations=ret;
 57   }
 58   else
 59      for(i=0;i<n;i++)
 60        if(de[i]>0)//如果存在环，不再往下算
 61         {
 62            flag=true;
 63            kind=2;
 64            relations=ret;
 65            break;
 66         }
 67     if(!flag)//如果都不满足，恢复
 68     {
 69         for(i=0;i<n;i++)
 70           de[i]=de2[i];
 71     }
 72     return;
 73 
 74 }
 75 int main()
 76 {
 77     char left,mid,right;
 78     int ni,nj;
 79     while(scanf("%d%d",&n,&m))
 80     {
 81         getchar();
 82         flag=false;
 83         kind=0;
 84         relations=0;
 85         if(n==0&&m==0)
 86          break;
 87         for(it=0;it<n;it++)//初始化
 88         {
 89             for(jt=0;jt<n;jt++)
 90               a[it][jt]=false;
 91             de[it]=-1;
 92             de2[it]=-1;
 93         }
 94         for(it=0;it<m;it++)
 95         {
 96             cin>>left>>mid>>right;
 97             getchar();
 98             ni=left-'A';//两点有联系时，入度先都置0
 99             nj=right-'A';
100             if(de[ni]==-1)
101             {
102                 de[ni]=0;
103                 de2[ni]=0;
104             }
105             if(de[nj]==-1)
106             {
107                 de[nj]=0;
108                 de2[nj]=0;
109             }
110             if(!a[ni][nj])
111             {
112                 a[ni][nj]=true;//ni和nj有关系
113                 de[nj]++;
114                 de2[nj]++;
115             }
116             if(!flag)
117             {
118                while(!Q1.empty())//没找到确定关系时，要一直恢复
119                  Q1.pop();
120                carry(n,it+1);
121             }
122         }
123         if(flag)
124         {
125             if(kind==1)
126             {
127                 printf("Sorted sequence determined after %d relations: ",relations);
128                 while(!Q1.empty())
129                 {
130                     cout<<Q1.front();
131                     Q1.pop();
132                 }
133                 cout<<"."<<endl;
134             }
135             else
136                 if(kind==2)
137                 {
138                   printf("Inconsistency found after %d relations.\n",relations);
139                 }
140         }
141         else
142             printf("Sorted sequence cannot be determined.\n");
143 
144     }
145 }

View Code

转载于:https://www.cnblogs.com/sdau--codeants/p/3389815.html