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.

有一些未知数各不相等，给出一些两两比较的大小关系，问到第几个关系开始可以确定整体排序或出现矛盾，再或者所有关系都用过了也无法确定整体排序
每输入一对关系，如果判定有结果，则可以忽略后面输入数据，即使后面输入数据能改变结果，也不用管。所以应该每输入一个关系就去更新当前的图，然后进行一趟拓扑排序。一旦产生结果，再对后面的数据处理下，就可以输出结果。 
（1）最终的图可以排序: 在输入结束前如果能得到最终的图（就是用这n个字母作为顶点，一个都不能少），而且最终得到的图无环只有唯一一个无前驱（即入度为0）的结点，但允许其子图有多个无前驱的结点。在这步输出排序后，不再对后续输入进行操作。
（2）输出矛盾: 在输入结束前如果最终图的子图有环,在这步输出矛盾后，不再对后续输入进行操作
（3）输出无法确认排序: 这种情况必须全部关系输入后才能确定，其中又有2种可能
    ①最终图的字母一个不缺，但是有多个无前驱结点
    ②输入结束了，但最终的图仍然字母不全，与无前驱结点的多少无关

#include <iostream>
#include <string.h>
using namespace std;
int n,m,map[27][27],deg[27],str[27];
int DFS()
{
    int temp[27]={0},k,c=0,flag=1;
    for(int i=0;i<n;i++)
        temp[i]=deg[i];
    for(int i=0;i<n;i++)
    {
        int count1=0;
        for(int j=0;j<n;j++)
            if(temp[j]==0)
            {
                count1++;
                k=j;
            }
        if(count1==0)return 0;//有环，有反向边
        if(count1>1)flag=-1;//有多个点的入度为0，成了一个无向图而不是一个有序的线性序列
        str[c++]=k;
        temp[k]=-1;
        for(int j=0;j<n;j++)
            if(map[k][j]==1)
            temp[j]--;
    }
    return flag;
}
int main()
{
    while(cin>>n>>m&&(n!=0&&m!=0))
    {
        memset(map,0,sizeof(map));
        memset(deg,0,sizeof(deg));
        memset(str,0,sizeof(str));
        char s[4];
        int flag=0;
        for(int i=1;i<=m;i++)
        {
            cin>>s;
            if(flag==1)continue;
            map[s[0]-'A'][s[2]-'A']=1;
            deg[s[2]-'A']++;
            int temp=DFS();
            if(temp==1)
            {
                cout<<"Sorted sequence determined after "<<i<<" relations: ";
                for(int j=0; j<n; j++)
                    cout<<char(str[j]+'A');
                cout<<"."<<endl;
                flag=1;
            }
            if(temp==0)
            {
                cout<<"Inconsistency found after "<<i<<" relations."<<endl;
                flag=1;
            }
        }
        if(flag==0)
            cout<<"Sorted sequence cannot be determined."<<endl;
    }
    return 0;
}