27-拓扑排序-poj1094

http://poj.org/problem?id=1094　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　

　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　Sorting It All Out

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 38379		Accepted: 13534

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

思路：就是利用拓扑排序，当条件不断增加时，依次判断，看是否能排好，同时要注意是否有环，注意在条件不足不能判断出先后顺序的情况下，依然要判断是否有环；

其实有一点有疑问，就是在前面条件充足，可以排好序之后即没有对后续的进行判断了，但是如果后面的条件使得前面的成环呢？貌似样例没有这个，可以直接ac

#include <iostream>
#include <cstring> 

using namespace std;
int in[30];  //入度 
int out[30]; //输出序列 
int map[30][30]; //有向图 

int TopoSort(int n){
	int ct = 0, cpin[30];	
	int flag = 1;
	for(int i = 0; i < n; i++){
		cpin[i] = in[i];
	}
	while(ct < n){
		int p = -1, m = 0;
		for(int i = 0; i < n; i++){
			if(cpin[i] == 0){
				p = i;
				m++;
			}
		}
		if(m == 0){
			return -1; //有环 
		}
		if(m > 1){
//			return 0;  //无法判断 
			flag = 0;  //得出不能判断只有还有检测，因为可能后面可能有环 
		}
		out[ct++] = p;
		cpin[p] = -1; 
		for(int i = 0; i < n; i++){
			if(map[p][i]){
				cpin[i]--;
	     	}
		}  
	}
	return flag;  //排好序了 
} 

int main(){
	std::ios::sync_with_stdio(false); 
	int n, m;
	char x, y;
	char ch;
	while(cin >> n >> m && (n != 0 && m != 0)){
		memset(in, 0, sizeof(in));
		memset(map, 0, sizeof(map));
		int flag = 1;
		for(int i = 0; i < m; i++){
			cin >> x >> ch >> y;
			if(flag == 0)
				continue;  //注意就算前面已经判断出结果，也要输入完，但不需要判断了 
			if(map[x - 'A'][y - 'A'] == 0){
				map[x - 'A'][y - 'A'] = 1;
				in[y - 'A']++;
				int s = TopoSort(n); 
				if(s == -1){
					cout << "Inconsistency found after ";
					cout << i + 1;
					cout << " relations."  << endl;
					flag = 0;
				} 
				if(s == 1){
					cout << "Sorted sequence determined after ";
					cout << i + 1;
					cout << " relations: ";
					for(int i = 0; i < n; i++){
						cout << char(out[i] + 'A');
					}
					cout << "." << endl;
					flag = 0;
				}
			}  
		}
		if(flag){
			cout << "Sorted sequence cannot be determined." << endl;
		}
	}
	return 0;
}