zoj 1060 Sorting It All Out 拓扑排序

//这题之前一直卡着没过，今晚蛋疼了一晚才过

//这题是拓扑排序，有几个易错点

//拓扑排序的时候，m每次要保证入度为0的点只有1个，因为题目要求的是完全排序

//当满足条件（确定排序或检测到矛盾）后，后面的输入就可以不管了

#include <iostream>

#include <queue>

#include <string>

#include <cstring>

using namespace std;



const int N = 30;



int maze[N][N];

bool isAppear[N];

int n, cases;

bool isDetermine, isInconsistance;

string path;



void toposort()

{

	queue<int> Q;

	int delete_num = 0;

	int cnt = 0;

	path = "";

	bool flag = false;

	int size[N];

	

	memset(size, 0, sizeof(size));

	for (int i = 0; i < n; i++)

	{		

		cnt = 0;

		for (int j = 0; j < n; j++)

			if (maze[j][i])

				size[i]++;

		if (!size[i])

			Q.push(i);

	}



	if (Q.empty())

	{

		isInconsistance = true;

		return;

	}



	if (Q.size() > 1)

		flag = true;

	

	while (!Q.empty())

	{

		int qhead = Q.front();

		Q.pop();

		path += char(qhead + 'A');

		delete_num++;

		for (int i = 0; i < n; i++)

			if (maze[qhead][i])

			{

				size[i]--;

				if (size[i] == 0)

					Q.push(i);

			}

		if (Q.size() > 1)

			flag = true;

	}

	

	cnt = 0;

	for (int i = 0; i < n; i++)

		if (isAppear[i])

			cnt++;



	if (cnt == delete_num && cnt == n && !flag)

		isDetermine = true;

	else if (cnt > delete_num)

		isInconsistance = true;

}



int main()

{

	char c1, c2;

	int num, ans;



	while (cin >> n >> cases && (n || cases))

	{

		memset(isAppear, false, sizeof(isAppear));

		isDetermine = false;

		isInconsistance = false;

		num = 0;

		memset(maze, 0, sizeof(maze));



		while (cases--)

		{

			num++;

			cin >> c1;

			cin >> c2;

			cin >> c2;

			if (isInconsistance || isDetermine)

				continue;



			isAppear[c1-'A'] = true;

			isAppear[c2-'A'] = true;

			maze[c1-'A'][c2-'A'] = 1;

			

			toposort();



			if (isDetermine)

				ans = num;

			if (isInconsistance)

				ans = num;

		}

		

		if (isDetermine || isInconsistance)

			if (isDetermine)

				cout << "Sorted sequence determined after " << ans << " relations: " << path << "." << endl;

			else

				cout << "Inconsistency found after " << ans << " relations." << endl;

		else

			cout << "Sorted sequence cannot be determined." << endl;

	}

	return 0;

}