POJ 1094 Sorting It All Out

题意：给你N个字母（ 输入时保证是字母表的前n个 ）， M个关系（ 关系全部为 X<Y 的格式 

 输出的答案有三种：

1.根据所给的关系可以确定唯一的拓扑排序（ 还要记录在给出的前几组就可以确定，不一定用到所有的关系 ）

2.所给的关系有相互矛盾的（同样要记录出根据前几组就可以得出这个结论）

3.用上所有的关系仍然不能确定出拓扑排序

首先，这道题用拓扑排序毋庸置疑，但是在排序的时候要做一些修改，题目的难点也就是在这里

具体详见代码：

 

                                                                                                                                           Sorting It All Out

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 23043		Accepted: 7965

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.

 

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
using namespace std;
#define MAXN 27

int in[MAXN], temp[MAXN], topo[MAXN]; //入度，topo排序结果
bool visit[MAXN][MAXN]; //标记i, j存在i < j关系
int n, m;//字母数，边数

int toposort() //拓扑排序，三个返回值：环、继续读边(未完成)、排序完成
{
	int u, v, count;
	int num; //入度为0的字母
	bool flag;
	count = 1; //已排序字母个数
	flag = true;
	memset(topo, 0, sizeof(topo));
	for(int i = 1; i <= n; ++i)
		temp[i] = in[i];
	for(int i = 1; i <= n; ++i)
	{
		num = 0; //这个不能写在外面，删边后度变化。。。。错了N久啊。。。
		for(int j = 1; j <= n; ++j) //统计入度为0的字母个数
			if(temp[j] == 0)
			{
				u = j;
				num++;
			}
		if(num == 0) //为环
			return 0;
		if(num > 1) //多条分支,继续读边
			flag = false;
		temp[u] = -1;
		topo[count++] = u; //加入拓扑序列
		for(int j = 1; j <= n; ++j) //删边操作
			if(visit[u][j] == true)
				temp[j]--;
	}
	if(flag == false)
		return -1;
	return 1; //经过n次排序到达这里，则排序完成
}

int main()
{
	char str[5];
	int start, end;
	bool flag;
	int res; //toposort返回的结果
	while(scanf("%d%d", &n, &m) != EOF)
	{
		if(n == 0 && m == 0)
			break;
		memset(in, 0, sizeof(in));
		memset(visit, false, sizeof(visit));
		flag = true;
		for(int i = 1; i <= m; ++i)
		{
			scanf("%s", str);
			if(flag == false) //出现环或者矛盾
				continue;
			start = str[0] - 'A' + 1;
			end = str[2] - 'A' + 1;
			in[end]++;
			visit[start][end] = true;
			res = toposort(); //加一次边拓扑排序一次
			if(res == 0)
			{
				printf("Inconsistency found after %d relations.\n", i);
				flag = false;
			}
			else if(res == 1)
			{
				printf("Sorted sequence determined after %d relations: ", i);
				for(int j = 1; j <= n; ++j)
					printf("%c", topo[j] + 'A' - 1);
				printf(".\n");
				flag = false;
			}
		}
		if(flag == true)
			printf("Sorted sequence cannot be determined.\n");
	}
	return 0;
}