题目链接：
啊哈哈，选我
题目：
Sorting It All Out
| Time Limit: 1000MS |
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Memory Limit: 10000K |
| Total Submissions: 26897 |
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Accepted: 9281 |
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
Sample Output
Sorted sequence determined after 4 relations: ABCD.
Inconsistency found after 2 relations.
Sorted sequence cannot be determined.
Source
East Central North America 2001
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这个题目对拓扑排序考虑的非常仔细。。
考虑了成环的情况。成环的情况也就是最后存在入度不为0的点。。则计数后最后的num不等于n。。
这个题目还有就是这个题目不是对所有的信息进行综合判断，而是根据前面的如果能够得到已经成环了，或者可以得到n的大小顺序了，则后面的就不用判断了。。。所以用ok1，ok2两个变量进行控制。。。
代码为：
#include
#include
#include
#include
#include
using namespace std; const int maxn=26+10; int n,m; int in[maxn],Copy[maxn],map[maxn][maxn],temp[maxn]; stack
S; int topo() { int flag=0,num=0; while(!S.empty()) S.pop(); memcpy(Copy,in,sizeof(in)); for(int i=0;i
1) flag=1; int Gery=S.top(); S.pop(); temp[num++]=Gery; for(int i=0;i