这道题目考察了拓扑排序的基本思想：每一步寻找一个入度为0的结点，然后 删除之。将这个结点指向的结点入度减1。删除从这个结点出发的所有边

同时考察了对于一个有向图是否有环、是否严格有序的判断。（当发现多个结点的度 为0时，则不是严格有序。当发现没有结点入度为0时，则有环） 需要注意的是逻辑上的关系。“Sorted sequence cannot be determined.”这个判断， 要放在最后，即只有发现了所有的点都没有环，且并没有发现严格有序性质的时候， 才可以判为“Sorted sequence cannot be determined."逻辑上的判决一定要注意优先级。

这个问题中优先级最高的是判断有环。一旦发现找不到度为0的结点，则立即return。

第二优先级是是否严格有序的判断。当每一步能且只能找到一个度为0的结点，则return。

第三优先级是无法确定。(当发现无法确定时，并不能立即return，因为还需要判断是否有环）

思维方式:当条件x成立时，另外的条件y，z是否一定不成立。如果一定不成立， 则可以断言return。否则必须继续判断下去。 提交记录： 1.样例未过！由于拓扑排序算法中，处理过程中入度减1的同时，并没有将对应的边删掉。 2.样例未过！ 由于在子函数judge中，对graph图进行了修改，导致后面处理失败。增加了memcpy解决。 4.wrong answer！ 在judge函数中，当出现多个结点度为0的时候，还需要继续判冲突，而不能直接return

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.

#include<cstdio>
#include<string.h>
#include<algorithm>
using namespace std;
int ans,m,n;
char top[50];
int s[50],t[50];//t数组记录v赢了几次,s数组保留赢的次数
int book[50][50];
int mark[30];
int tuopu()
{int l=0,flag=0;for(int h=1; h<=ans; h++){int k=-1,num=0;for(int i=1; i<=n; i++){if(mark[i]&&s[i]==0)//字母出现过,但是一次也没赢{k=i;num++;}}if(k==-1)return -1;if(num!=1)flag=1;s[k]=-1;top[l]='A'+(k-1);l++;for(int i=1; i<=n; i++){if(book[k][i])s[i]--;}}top[l]=0;if(ans==n&&flag==0)return 1;return 0;
}
int main()
{while(~scanf("%d%d",&n,&m)&&(n+m))//n个字母,m组关系{ans=0;int flag=0;char c[10];memset(t,0,sizeof(t));memset(mark,0,sizeof(mark));//标记字母是否出现过memset(book,0,sizeof(book));for(int i=1; i<=m; i++){scanf("%s",c);if(flag==1)//已找到目标情况continue;int u=c[0]-'A'+1;int v=c[2]-'A'+1;if(mark[u]==0){mark[u]=1;ans++;}if(mark[v]==0){mark[v]=1;ans++;}if(book[u][v]==0){t[v]++;book[u][v]=1;for(int j=0; j<30; j++)s[j]=t[j];//将赢的次数保留下来int a=tuopu();if(a==-1)//有环{printf("Inconsistency found after %d relations.\n",i);flag=1;}else if(a==1)//有序{printf("Sorted sequence determined after %d relations: %s.\n",i,top);flag=1;}}}if(flag==0)printf("Sorted sequence cannot be determined.\n");}return 0;
}