nyoj 349 (poj 1094) (拓扑排序)

Sorting It All Out

时间限制：3000 ms  |  内存限制：65535 KB

难度：3

 

描述

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.

 

输入

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.

输出

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.

样例输入

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

样例输出

Sorted sequence determined after 4 relations: ABCD.
Inconsistency found after 2 relations.
Sorted sequence cannot be determined.

/**
    拓扑排序实现步骤：
        ①、在有向图中找到没有前驱节点的顶点并输出
        ②、删除该点和该点所到的下一个点之间的线
        ③、重复①、②操作
**/

 /**
     分析:
         Ⅰ、数据 n 表示有 n 个字母， 且字母由A开始到 char('A' + n - 1)
         Ⅱ、 数据 m 表示有 m 个输入， 用来判断是否能够确定前 n 个字符的关系
         Ⅲ、如果确定其前 n 个字母的先后顺序就输出，同时对后面的数据不做判断
         Ⅳ、如果能够确定有环的存在也可以不用对后面的数据进行判断 （输出 Inconsistency...）
 **/

核心代码：

 int topo_sort () {
     int flag = , temp [], c = , Q[], in_num, pos;
     for (int i = ; i <= n; ++ i) {
         temp [i] = my_in [i];
     }
     for (int i = ; i <= n; ++ i) {
         in_num = ;
         for (int j = ; j <= n; ++ j) {
             if (!temp [j]) {
                 pos = j;
                 in_num ++;
             }
         }
         if (!in_num) return ; // 环
         if (in_num > ) flag = -; // 不可能有序，但有可能有环，所以不能return

         Q [c ++] = pos;
         temp [pos] = -;
         for (int j = ; j <= n; ++ j) {
             if (my_map [pos][j] == ) {
                 temp [j] --;
             }
         }
     }
     return flag;
 }

C/C++代码实现（AC）:

 #include <iostream>
 #include <algorithm>
 #include <cstdio>
 #include <cstring>

 using namespace std;

 int n,     m, my_map [][], my_in [], Q[], c;

 int topo_sort () {
     int flag = , temp [], num_in, pos;
     c = ;
     for (int i = ; i <= n; ++ i) {
         temp [i] = my_in [i];
     }

     for (int i = ; i <= n; ++ i) {
         num_in = ;
         for (int j = ; j <= n; ++ j) {
             if (!temp [j]) {
                 ++ num_in;
                 pos = j;
             }
         }
         if (!num_in) return ; // 环
         if (num_in > ) flag = -; // 不能确定

         temp [pos] = -;
         Q [c ++] = pos;
         for (int j = ; j <= n; ++ j) {
             if (my_map [pos][j] == ) {
                 temp [j] --;
             }
         }
     }
     return flag;
 }

 int main () {
     while (scanf ("%d%d", &n, &m), n !=  || m != ) {
         int flag = , a1, b1;
         char a, b, d;
         memset (my_map, , sizeof (my_map));
         memset (my_in, , sizeof (my_in));

         for (int i = ; i <= m; ++ i) {
             getchar ();
             scanf ("%c%c%c", &a, &d, &b);
             if (!flag) continue;

             a1 = int (a - 'A' + );
             b1 = int (b - 'A' + );
             my_map [a1][b1] = ;
             my_in [b1] ++;

             int x = topo_sort ();
             if (x == ) {
                 printf("Sorted sequence determined after %d relations: ",i);
                 for (int i = ; i < c; ++ i) {
                     printf ("%c", 'A' + Q [i] - );
                 }
                 printf (".\n");
                 flag = ;
             } else if (x == ) {
                 printf("Inconsistency found after %d relations.\n",i);
                 flag = ;
             }
         }

         if (flag) {
             printf("Sorted sequence cannot be determined.\n");
         }
     }
 }