Sorting It All Out (拓扑排序+思维)-白红宇

Sorting It All Out (拓扑排序+思维)

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发布时间：2019-06-12

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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 6A

Sample Output

Sorted sequence determined after 4 relations: ABCD.Inconsistency found after 2 relations.Sorted sequence cannot be determined. 代码：

#include
     
      #include
      
       #include
       
        #include
        
         #include
         
          #include
          
           #include
           
            #include
            #include
             
              const int maxn=1e5+5;typedef long long ll;using namespace std;vector
              
               vec[30];int n,m;int du[30];int chu[30];int du1[30];int flag;vector
               
                ans1;void Tpsort(){ priority_queue
                
                 
                  ,greater
                  
                    >q; priority_queue
                   
                    q1; int s=0; for(int t=0;t

ans,ans2; while(!q.empty()) { int now=q.top(); int now2=q1.top(); q.pop(); q1.pop(); ans.push_back(now); ans2.push_back(now2); for(int t=0;t

>n>>m) { if(n==0&&m==0) { break; } for(int t=0;t

转载于:https://www.cnblogs.com/Staceyacm/p/11261298.html

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