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            POJ 1094 Sorting It All Out 拓撲排序

            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.

            Source

                拓撲排序算法:          1.將所有入度為0的點加入隊列;
                                            2.彈出隊首元素u,輸出u并將所有與u關聯的頂點v的入度減1;如果v的入度為0,將u加入隊列;
                                            3.重復第2步,如果所有的頂點都被訪問到,則輸出序列是一個拓撲排序;否則該DAG圖中存在環路。
            #include <iostream>
            #include 
            <string>
            #include 
            <vector>
            #include 
            <queue>
            using namespace std;

            int n,m;
            vector
            <int> top;
            vector
            <int> in;
            vector
            < vector<int> > map;

            int topsort(){
                
            int i,u;
                
            bool flag=false;
                queue
            <int> q;
                vector
            <int> d(in.begin(),in.end());
                
            for(i=0;i<n;i++)
                    
            if(!d[i]) q.push(i);
                top.clear();
                
            while(!q.empty()){
                    
            if(q.size()!=1) flag=true;
                    u
            =q.front();
                    q.pop();
                    top.push_back(u);
                    
            for(i=0;i<map[u].size();i++)
                        
            if(--d[map[u][i]]==0) q.push(map[u][i]);
                }

                
            if(top.size()!=n) return 1;
                
            if(flag) return 0;
                
            return 2;
            }

            int main(){
                
            string str;
                
            int i,j,u,v,ans;
                
            while(cin>>n>>m,n||m){
                    
            in.assign(n,0);
                    map.assign(n,vector
            <int>());
                    
            for(ans=i=0;i<&& !ans;i++){
                        cin
            >>str;
                        u
            =str[0]-'A',v=str[2]-'A';
                        
            if(find(map[u].begin(),map[u].end(),v)==map[u].end())
                            map[u].push_back(v),
            in[v]++;
                        ans
            =topsort();
                    }

                    
            for(j=i;j<m;j++) cin>>str;
                    
            switch(ans){
                        
            case 0:cout<<"Sorted sequence cannot be determined."<<endl;break;
                        
            case 1:cout<<"Inconsistency found after "<<i<<" relations."<<endl;break;
                        
            case 2:{
                            cout
            <<"Sorted sequence determined after "<<i<<" relations: ";
                            
            for(j=0;j<n;j++) cout<<char('A'+top[j]);
                            cout
            <<"."<<endl;
                            
            break;
                               }

                    }

                }

                
            return 0;
            }

            posted on 2009-05-19 20:43 極限定律 閱讀(629) 評論(0)  編輯 收藏 引用 所屬分類: ACM/ICPC

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