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  • Sorting It All Out (拓扑排序+思维)

    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.
    代码:
    #include<cstdio>
    #include<iostream>
    #include<cstring>
    #include<algorithm>
    #include<queue>
    #include<stack>
    #include<vector>
    #include<map>
    #include<cmath>
    const int maxn=1e5+5;
    typedef long long ll;
    using namespace std;
    vector<int>vec[30];
    int n,m;
    int du[30];
    int chu[30];
    int du1[30];
    int flag;
    vector<int>ans1;
    void Tpsort()
    {
        priority_queue<int,vector<int>,greater<int> >q;
        priority_queue<int>q1;
        
        int s=0;
        for(int t=0;t<n;t++)
        {
            if(chu[t]==0&&du[t]==0)
            {
             s=1;    
            }
        }
        for(int t=0;t<n;t++)
        {
            du1[t]=du[t];
        }
        for(int t=0;t<n;t++)
        {
            if(du1[t]==0)
            {
                q.push(t);
                q1.push(t);
            }
        }
        vector<int>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<vec[now].size();t++)
            {
                int next=vec[now][t];
                du1[next]--;
                if(du1[next]==0)
                {
                    q.push(next);
                    q1.push(next);
                }
            }
        }
        if(ans.size()!=n)
        {
            flag=1;
        }
    //    cout<<ans.size()<<" "<<s<<endl;
         
        if(ans.size()==n&&s==0)
        {
            int sss=0;
            for(int t=0;t<ans.size();t++)
            {
                if(ans[t]!=ans2[t])
                {
                    sss=1;
                }
            }
            if(sss==0)
            {
                flag=2;
                for(int t=0;t<ans.size();t++)
                {
                    ans1.push_back(ans[t]);
                }
            }
            
            
        }
        
    }
    int main()
    { 
      
       while(cin>>n>>m)
       {
        if(n==0&&m==0)
        {
            break;
        }
        for(int t=0;t<n;t++)
        {
            vec[t].clear();
        }
           char str[5];
           memset(du,0,sizeof(du));
           memset(chu,0,sizeof(chu));
           flag=0;
           int ss=0;
           int k;
           ans1.clear();
           for(int t=1;t<=m;t++)
           {
               scanf("%s",str);
               vec[str[0]-'A'].push_back(str[2]-'A');
               du[str[2]-'A']++;
               chu[str[0]-'A']++;
               if(ss)
               {
               continue;    
            }
               Tpsort();
               if(flag==1)
               {
                   printf("Inconsistency found after %d relations.
    ",t); 
                   ss=1;
            }
            else if(flag==2)
            {
                ss=1;
                printf("Sorted sequence determined after %d relations: ",t); 
                for(int j=0;j<n;j++)
                {
                    printf("%c",ans1[j]+'A');
                }
                printf(".
    ");
                
            }
        }
        if(ss==0)
        puts("Sorted sequence cannot be determined.");
        
       }
       return 0;
    }
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  • 原文地址:https://www.cnblogs.com/Staceyacm/p/11261298.html
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