[题解] poj 1094 Sorting It All Out (拓扑排序)

- 传送门 -

　http://poj.org/problem?id=1094

#Sorting It All Out

| Time Limit: 1000MS |   | Memory Limit: 10000K |
| Total Submissions: 35759 |   | Accepted: 12590 |

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.

Source

East Central North America 2001

- 题意 -

　有 n 个大写字母 (A 到 A + n - 1), m 组大小关系.
　设到第 k 组的时候能确定唯一的大小关系或确定发生了冲突, 求这个 k 值, 或是输完 m 组也没有冲突, 也不能确定唯一顺序.
　

- 思路 -

　好模板哦!
　注意一下三种情况的判断顺序就好.
　
　细节见代码.
　

- 代码 -

#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;

const int N = 30;

int ANS[N], TMP[N], IN[N], MAP[N][N];
int n, m, ans, sz, f;
char S[10];
queue<int> q;

int topo() {
	sz = 0, f = 0;
	while (!q.empty())
		q.pop();
	for (int i = 1; i <= n; ++i)
		if (TMP[i] == 0) {
			q.push(i);
			TMP[i]--;
		}
	while (!q.empty()) {
		if (q.size() > 1) f = -1;
		int i = q.front();
		q.pop();
		ANS[++sz] = i;
		for (int j = 1; j <= n; ++j) {
			if (MAP[i][j]) TMP[j]--;
			if (TMP[j] == 0) {
				q.push(j);
				TMP[j]--;
			}
		}
	}
	for (int i = 1; i <= n; ++i)
		if (TMP[i] != -1)
			return -1;
	if (f == -1) return 0;
	return 1; //注意顺序, 先判环的存在, 再判无法确定顺序的情况.
}

int main() {
	while(~scanf("%d%d", &n, &m) && n && m) {
		memset(IN, 0, sizeof (IN));
		memset(MAP, 0, sizeof (MAP));
		ans = 0;
		for (int i = 1; i <= m; ++i) {
			scanf("%s", S);
			if (ans != 0) continue;
			int a = S[0] - 'A' + 1, b = S[2] - 'A' + 1;
			if (!MAP[b][a]) IN[a] ++;
				MAP[b][a] = 1;
			for (int j = 1; j <= n; ++j)
				TMP[j] = IN[j];
			ans = topo();
			if (ans == 1) {
				printf("Sorted sequence determined after %d relations: ", i);
				for (int j = n; j >= 1; --j)
					printf("%c", ANS[j] + 'A' - 1);
				printf(".
");
			}
			if (ans == -1)
				printf("Inconsistency found after %d relations.
", i);
		}
		if (ans == 0)
			printf("Sorted sequence cannot be determined.
");
	}
	return 0;
}