POJ-1094 Sorting it all out (拓扑排序)

Sorting It All Out

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 26070		Accepted: 9031

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.
 
  

 
  
思路：
 
  
拓扑排序（堆栈）；
 
  
需要记录的数据： 有向图（临界矩阵），  各点入度值，排序结果
 
  
先找入度为0的入栈，作为起点，放到栈顶；
 
  
再将与栈顶元素相连的点的入度减1、如果减1后有点入度为0，则入队；
 
  
类推，详见代码
 
  

 
  
代码：
 
  
 
  
#include 
#include 
#include 
#define N 30

using namespace std;

int n, m;
bool map[N][N];
int in[N], list[N];

int toposort()
{
	int tin[N];
	memcpy(tin, in, sizeof(in));
	stackq;

	for(int i = 0; i < n; i ++){		// 先找一个入度为1的点
		if(!tin[i]){
			q.push(i);
		}
	}
	
	int flag = 0, top, l_ct = 0;		// flag 为路径唯一的标记， l_ct 为 list 数组的计数器
	while(!q.empty()){
		if(q.size() > 1)
			flag = 1;

		top = q.top();
		q.pop();

		list[l_ct ++] = top;

		for(int i = 0; i < n; i ++){
			if(map[top][i]){
				if(-- tin[i] == 0){		// 入度为1的点入栈
					q.push(i);
				}
			}
		}
	}

	if(l_ct != n){			// 不能拓扑排序，既有环（必须先判断此项）
		return 1;
	}
	else if(flag == 1){		// 有多种排序方式
		return 2;
	}

	return 0;					// 成功
}

int main()
{
	char x, y;
	while(scanf("%d%d", &n, &m), n && m){
		int failed = 0, sure = 0;		// 已失败（有环）， 已成功排序
		memset(map, 0, sizeof(map));
		memset(in, 0, sizeof(in));
		memset(list, 0, sizeof(list));

		for(int i = 1; i <= m; i ++){
			scanf(" %c<%c", &x, &y);

			if(!failed && !sure){
			    if(map[y - 'A'][x - 'A']){
					printf("Inconsistency found after %d relations.\n",i);
					failed = 1;
					continue;
				}
		
				if(map[x - 'A'][y - 'A'] == 0){
					map[x - 'A'][y - 'A'] = 1;	
					in[y - 'A'] ++;					// y点入度加1
				}

				int ans = toposort();

				if(ans == 0){
					printf("Sorted sequence determined after %d relations: ",i);
					for(int j = 0; j < n; j ++)
						printf("%c", 'A' + list[j]);
					printf(".\n");
					sure = 1;
				}
				else if(ans == 1){
					printf("Inconsistency found after %d relations.\n",i);
					failed = 1;
				}
			}
		}
		if(!failed && !sure){				// 即未成功也未失败，既无法判断
			printf("Sorted sequence cannot be determined.\n");
		}
	}

	return 0;
}