POJ 1094 Sorting It All Out（入度法求拓扑排序，成环判断）

题目来源：http://poj.org/problem?id=1094

问题描述

Sorting It All Out

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 39161		Accepted: 13808

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

------------------------------------------------------------

题意

给定一组字母间两两顺序关系，问是否能确定这些字母的唯一顺序。如果能确定，输出前几条关系就能确定；如果矛盾，输出前几条关系处发生矛盾；如果不能确定，输出不能确定。

一些题目中没有说但很重要的约定：

1. 如果前x条关系就能确定唯一顺序，但第y条（y>x）关系开始又产生了矛盾，此时按“前x条关系能确定唯一顺序”处理；

2. 如果既发生矛盾又不能确定关系，优先输出矛盾。

------------------------------------------------------------

思路

本题需要判断是否成环，用基于入度的拓扑排序算法。

1. 找到图中入度为0的点作为拓扑排序的起点（忽略孤立点，若有孤立点则一定不能确定唯一顺序，但仍有可能是有环）

2. 将该点的所有后继的入度-1（相当于删除该节点）

3. 循环“步骤1-步骤2”n次，某次循环如果不存在入度为0的点则说明有环（“矛盾”），直接退出循环；如果存在不止一个入度为0且不孤立的点，则一定不能确定唯一顺序，但仍有可能有环

------------------------------------------------------------

代码

// 拓扑排序：基于入度的实现
// 找到图中入度为0的点作为拓扑排序的起点
// 将该点的后继节点的入度-1（相当于删除该点），再遍历全图找入度为0的点
// 如此循环
// 注意：
// 1. 优先判断是否成环，再判断是否可以确定拓扑顺序
// 2. 如果已经可以确定拓扑顺序，即使后面新增的语句的加入使得图成环，也不管了

#include
#include
#include

const int NMAX = 30;
std::vector E[NMAX];				// 邻接表
int indegree[NMAX];						// 入度数组
bool vis[NMAX];							// 是否访问数组

void clear_case()						// 每算完一个测试用例清空邻接表和入度数组
{
	for (int i = 0; i < NMAX; i++)
	{
		E[i].clear();
	}
	memset(indegree, 0, sizeof(indegree));
	memset(vis, 0, sizeof(vis));
}

int TopoSort(int n, std::vector &res)	// res: 拓扑排序结果数组
{
	res.clear();							// 清空结果数组
	memset(indegree, 0, sizeof(indegree));	// 清空入度数组
	memset(vis, 0, sizeof(vis));			// 清空访问标记
	int i, j, k, u, ucnt = 0, ans = 0;
	for (i=0; i 1)				// 入度为0的节点多于1个
		{
			ans = 1;					// 不能确定拓扑关系，但可能有环，不能返回，要继续计算
		}
		vis[u] = 1;
		res.push_back(u);
		if (E[u].empty())
		{
			continue;
		}
		for (j=0; j res;
	std::vector::iterator it;
	while (scanf("%d%d", &n, &k))
	{
		if (n==0 && k==0)
		{
			break;
		}
		has_print = false;
		clear_case();								//	清空全局变量
		for (i=0; i