POJ-1094 Sorting It All Out -----拓扑排序判断状态

Sorting It All Out

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 21865		Accepted: 7529

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.

/*
这道题WA了好久,其中有几个需要注意的地方
1、当出现正好存在一种情况能够排序完所有节点时，不管以后的边会出现什么情况，都输出
    能够排序成功
2、当中间的拓扑排序过程中出现多个几点的入度为0时，只记录当时的状态
    （亦即该测试数据要么出现环，要么就是有多组解），不能立即返回，
    要继续读边，直到能够排序完成（此时输出有多解的情况）或者出现环。
*/
#include<cstdio>
#include<iostream>
#include<cstring>

using namespace std;

bool G[30][30];
int d[30];
char s[30];

int toposort(int n)
{
	int num,k,i,j,t;
	int td[30];
	bool flag=true;
	for(i=1;i<=n;++i)
		td[i]=d[i];
	memset(s,0,sizeof(0));
	for(j=0;j<n;++j)
	{
		num=0;
		for(i=1;i<=n;++i)
		{
			if(td[i]==0)
			{
				k=i;
				++num;
			}
		}
		if(num==0)    //有环
			return -1;
		if(num>1)  //有多种情况，还需继续读边判断
        {
            flag=false;
        }
		s[j]='A'+k-1;
		td[k]=-1;
		for(t=1;t<=n;++t)
		{
			if(G[k][t])
				--td[t];
		}
	}
	s[n]='\0';
	if(flag==false)  //情况不唯一
        return 0;
	return 1;      //全部排好序了返回1.
}

int main()
{
	int n,m,i,ans,k;
	bool flag;
	char temp[5];
	while(scanf("%d%d",&n,&m),m||n)
	{
		memset(G,false,sizeof(G));
		memset(d,0,sizeof(d));
		flag=true;
		for(i=1;i<=m;++i)
		{
			scanf("%s",temp);
			if(!flag)         //已经排好序或者有环
				continue;
			int u=temp[0]-'A'+1;
			int v=temp[2]-'A'+1;
			if(!G[u][v])
			{
				++d[v];
				G[u][v]=true;
			}
			ans=toposort(n);
			if(ans==1||ans==-1)
			{
				k=i;
				flag=false;
			}
		}
		if(ans==1)
		{
			printf("Sorted sequence determined after %d relations: %s.\n",k,s);
		}
		else if(ans==-1)
		{
			printf("Inconsistency found after %d relations.\n", k);
		}
		if(flag&&i>m)
			printf("Sorted sequence cannot be determined.\n");
	}
	return 0;
}