POJ 1094 Sorting It All Out 拓扑排序

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

    拓扑排序算法:          1.将所有入度为0的点加入队列;
                                2.弹出队首元素u,输出u并将所有与u关联的顶点v的入度减1;如果v的入度为0,将u加入队列;
                                3.重复第2步,如果所有的顶点都被访问到,则输出序列是一个拓扑排序;否则该DAG图中存在环路。
#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 极限定律 阅读(611) 评论(0)  编辑 收藏 引用 所属分类: ACM/ICPC


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