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 = 1, temp [30], c = 0, Q[30], in_num, pos;
    for (int i = 1; i <= n; ++ i) {
        temp [i] = my_in [i];
    }
    for (int i = 1; i <= n; ++ i) {
        in_num = 0;
        for (int j = 1; j <= n; ++ j) {
            if (!temp [j]) {
                pos = j;
                in_num ++;
            }
        }
        if (!in_num) return 0; // 环
        if (in_num > 1) flag = -1; // 不可能有序，但有可能有环，所以不能return
        
        Q [c ++] = pos;
        temp [pos] = -1;
        for (int j = 1; j <= n; ++ j) {
            if (my_map [pos][j] == 1) {
                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 [30][30], my_in [30], Q[30], c;

int topo_sort () {
    int flag = 1, temp [30], num_in, pos;
    c = 0;
    for (int i = 1; i <= n; ++ i) {
        temp [i] = my_in [i];
    }

    for (int i = 1; i <= n; ++ i) {
        num_in = 0;
        for (int j = 1; j <= n; ++ j) {
            if (!temp [j]) {
                ++ num_in;
                pos = j;
            }
        }
        if (!num_in) return 0; // 环
        if (num_in > 1) flag = -1; // 不能确定

        temp [pos] = -1;
        Q [c ++] = pos;
        for (int j = 1; j <= n; ++ j) {
            if (my_map [pos][j] == 1) {
                temp [j] --;
            }
        }
    }
    return flag;
}

int main () {
    while (scanf ("%d%d", &n, &m), n != 0 || m != 0) {
        int flag = 1, a1, b1;
        char a, b, d;
        memset (my_map, 0, sizeof (my_map));
        memset (my_in, 0, sizeof (my_in));

        for (int i = 1; i <= m; ++ i) {
            getchar ();
            scanf ("%c%c%c", &a, &d, &b);
            if (!flag) continue;

            a1 = int (a - 'A' + 1);
            b1 = int (b - 'A' + 1);
            my_map [a1][b1] = 1;
            my_in [b1] ++;

            int x = topo_sort ();
            if (x == 1) {
                printf("Sorted sequence determined after %d relations: ",i);
                for (int i = 0; i < c; ++ i) {
                    printf ("%c", 'A' + Q [i] - 1);
                }
                printf (".
");
                flag = 0;
            } else if (x == 0) {
                printf("Inconsistency found after %d relations.
",i);
                flag = 0;
            }
        }

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