Sorting It All Out

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.

題意：給定一組字母的大小關係判斷它們是否能組成唯一的拓撲序列。
思路：

1. 先判斷是否有環（一旦發現找不到入度為0的結點，立即return）

2. 再判斷是否嚴格有序（當每一步能且只能找到一個度為0的結點，則輸出）

3. 最後才能判斷是否能得出結果（當發現無法確定時，並不能立即return，因為還需要判斷是否有環）。

當發現多個結點的度為0時，則不是嚴格有序。當發現沒有結點入度為0時，則有環
必須遍歷完整個圖才能知道是否有環!!!

注意輸出中的標點符號

#include<cstdio>
#include<cstring>
using namespace
 
 std;
int n,m,ok;
int map[30][30];//鄰接矩陣存圖
int deg[30];//入度
char s[10];
int q[30];//模擬佇列儲存要輸出的序列
int TP(){
	int cnt=0;//佇列中最後一個元素的下標（解空間中零入度頂點的個數）
	int temp[30];//對頂點入度備份
	int pos;//記錄一個零入度頂點位置
	int m;//零入度頂點的個數
	int flag=1;//flag==1有序  flag==-1不能確定
	for(int i=1;i<=n;i++){
		temp[i]=deg[i];//備份 
	} 
	for(int i=1;i<=n;i++){//遍歷n遍，必須把圖全部遍歷完
		m=0;
		for(int j=1;j<=n;j++){//查詢零入度頂點的個數
			if(temp[j]==0){
				m++;
				pos=j;//記錄一個零入度頂點的位置
			}
		}
		if(m==0){       //有環 
			return 0;
		}
		if(m>1){//無序
			flag=-1;//不能立即退出，還要繼續遍歷判斷是否有環
		} 
		q[cnt++]=pos;//零入度頂點入度
		temp[pos]=-1;//將零入度頂點的入度置為-1
		for(int j=1;j<=n;j++){//刪除已pos為起點的邊
			if(map[pos][j]==1)
			temp[j]--;//相應頂點的入度減1；
		}
	}
	return flag;
}
int main(){
	while(scanf("%d%d",&n,&m)){
		if(n==0&&m==0) break;
		memset(map,0,sizeof(map));
		memset(deg,0,sizeof(deg));
		ok=0;
		for(int i=1;i<=m;i++){
			scanf("%s",s);
			if(ok) continue;//ok!=0時，已得出結果，將不考慮後面輸入
			int u=s[0]-'A'+1;
			int v=s[2]-'A'+1;
			map[u][v]=1;
			deg[v]++;
			int flag=TP();
			if(flag==0){//有環
				printf("Inconsistency found after %d relations.\n",i);//注意標點符號
				ok=1;
			}
			else if(flag==1){//有序
				printf("Sorted sequence determined after %d relations: ",i);
				for(int j=0;j<n;j++){
					printf("%c",q[j]+'A'-1);
				}
				printf(".\n");//注意標點符號
				ok=1;
			}
		}
		if(!ok){//無法得出結果
			printf("Sorted sequence cannot be determined.\n");//注意標點符號
		}	
	}
	return 0;
}

注意輸出中的標點符號