poj.1094.Sorting It All Out（topo)

Sorting It All Out

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 28762		Accepted: 9964

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

East Central North America 2001

 #include<stdio.h>
 #include<string.h>
 #include<queue>
 using namespace std;
 int n , m ;
 bool map[][] ;
 int in[] ;
 char st[][] ;
 char a[] ;

 int topo ()
 {
     queue <int> q ;
     int indegree[] ;
     int cnt =  ;
     int k =  ;
     int ans = - ;//record the condition that "cnt > 1"
     while (!q.empty ())
         q.pop () ;
     for (int i =  ; i <= n ; i++)
         indegree[i] = in[i] ;
     for (int i =  ; i <= n ; i++) {
         if (in[i] == ) {
             q.push (i) ;
             cnt++ ;
         }
       //  printf ("%d " , in[i]) ;
     }
     if (cnt > )
         ans =  ;//conditons are not satisfied
     while (!q.empty ()) {
         int temp = q.front () ;
         a[k++] = 'A' + temp -  ;
         q.pop () ;
         cnt =  ;
         for (int i =  ; i <= n ; i++) {
             if (map[temp][i]) {
                 indegree[i]-- ;
                 if (indegree[i] == ) {
                     cnt++ ;
                     q.push (i);
                 }
             }
         }
         if (cnt > )
             ans =  ;//conditons are not satisfied
     }
     if (k != n )
         return  ;//there is a circle
     if (k == n && ans != )
         return   ;//success

     if (ans == )
         return  ;
 }

 int main ()
 {
    // freopen ("a.txt" , "r" , stdin) ;
     int link ;
     int flag ;
     int success ;
     while (~ scanf ("%d%d" , &n , &m)) {
         if (n ==  && m ==)
             break ;
         getchar () ;
         memset (in ,  , sizeof(in)) ;
         memset (map ,  , sizeof(map)) ;
         link = - ;
         flag = - ;
         success = - ;
         for (int i =  ; i < m ; i++)
             gets (st[i]) ;

         for (int i =  ; i < m ; i++) {
             int u = st[i][] - 'A' +  ;
             int v = st[i][] - 'A' +  ;
          //   printf ("u = %d , v = %d\n" , u , v) ;
             if (!map[u][v])
                 in[v]++ ;
             map[u][v] =  ;

             flag = topo () ;
             if (flag == ) {
                 link = i +  ;
                 break ;
             }
             else if(flag == ) {
                 success = i +  ;
                 break ;
             }
          //   puts ("") ;
         }
         if (flag == ) {
             printf ("Inconsistency found after %d relations.\n" , link) ;
         }
         else if (flag == ) {
             puts ("Sorted sequence cannot be determined.") ;
         }
         else {
             printf ("Sorted sequence determined after %d relations: " , success) ;
             for (int i =  ; i < n ; i++) {
                 printf ("%c" , a[i]) ;
             }
             printf (".\n") ;
         }
     }
     return  ;
 }

要一条条边测下来，not determined 肯定是最后一条边出来才能 判断 ，但在这之前若 先判断出了 success 或 inconsistency 就能结束了（一开始要把所有边先记录下来）

success只有 无环 & 同一时刻加入队列的点 == 1 时才能 成立