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[30][30] ;
int in[30] ;
char st[1000][5] ;
char a[30] ;

int topo ()
{
    queue <int> q ;
    int indegree[30] ;
    int cnt = 0 ;
    int k = 0 ;
    int ans = -1 ;//record the condition that "cnt > 1"
    while (!q.empty ())
        q.pop () ;
    for (int i = 1 ; i <= n ; i++)
        indegree[i] = in[i] ;
    for (int i = 1 ; i <= n ; i++) {
        if (in[i] == 0) {
            q.push (i) ;
            cnt++ ;
        }
      //  printf ("%d " , in[i]) ;
    }
    if (cnt > 1)
        ans = 1 ;//conditons are not satisfied
    while (!q.empty ()) {
        int temp = q.front () ;
        a[k++] = 'A' + temp - 1 ;
        q.pop () ;
        cnt = 0 ;
        for (int i = 1 ; i <= n ; i++) {
            if (map[temp][i]) {
                indegree[i]-- ;
                if (indegree[i] == 0) {
                    cnt++ ;
                    q.push (i);
                }
            }
        }
        if (cnt > 1)
            ans = 1 ;//conditons are not satisfied
    }
    if (k != n )
        return 0 ;//there is a circle
    if (k == n && ans != 1)
        return  2 ;//success

    if (ans == 1)
        return 1 ;
}

int main ()
{
   // freopen ("a.txt" , "r" , stdin) ;
    int link ;
    int flag ;
    int success ;
    while (~ scanf ("%d%d" , &n , &m)) {
        if (n == 0 && m ==0)
            break ;
        getchar () ;
        memset (in , 0 , sizeof(in)) ;
        memset (map , 0 , sizeof(map)) ;
        link = -1 ;
        flag = -1 ;
        success = -1 ;
        for (int i = 0 ; i < m ; i++)
            gets (st[i]) ;

        for (int i = 0 ; i < m ; i++) {
            int u = st[i][0] - 'A' + 1 ;
            int v = st[i][2] - 'A' + 1 ;
         //   printf ("u = %d , v = %d\n" , u , v) ;
            if (!map[u][v])
                in[v]++ ;
            map[u][v] = 1 ;

            flag = topo () ;
            if (flag == 0) {
                link = i + 1 ;
                break ;
            }
            else if(flag == 2) {
                success = i + 1 ;
                break ;
            }
         //   puts ("") ;
        }
        if (flag == 0) {
            printf ("Inconsistency found after %d relations.\n" , link) ;
        }
        else if (flag == 1) {
            puts ("Sorted sequence cannot be determined.") ;
        }
        else {
            printf ("Sorted sequence determined after %d relations: " , success) ;
            for (int i = 0 ; i < n ; i++) {
                printf ("%c" , a[i]) ;
            }
            printf (".\n") ;
        }
    }
    return 0 ;
}

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