POJ 1094：Sorting It All Out（拓撲排序+傳遞閉包）

Sorting It All Out

Time Limit: 1000MS	Memory Limit: 10000K
Total Submissions: 19834	Accepted: 6784

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

對前n個字母，輸入m組确定他們之間的大小關系，進而判斷有沒有唯一的序列。      

第一種輸出，在前t(t<=m)已能确定一個序列      

第二種:不能确定唯一的序列      

第三種：不存在拓撲排序      

思路：拓撲排序+floyd。先用拓撲排序判斷在目前情況下，有沒有環的存在，如果有環的存在則證明m組關系是有沖突的，      

即可輸出3。若沒有環，那沒接下來就是判斷能否構成唯一的有序序列，用floyd傳遞閉包，如果能構成唯一的有序序列，      

則頂點1~n，必分布着0~n-1個入度，才能保證0個入度的頂點排在第1位，1個入度的頂點排在第2位（因為隻有1個頂點比它大），      

以此類推。有一點要注意下，就是要每輸入一次判斷一下，如果有輸出了，以後就隻輸入就好，不用第二次輸出。      

源代碼：      

#include<iostream>
using namespace std;

bool graph[30][30];
int n,m;
int path[30];

bool TopSort()
{
	int i,j,k;
	int indegree[30];
	for(i=0;i<n;i++)
	{
		indegree[i]=0;
		for(j=0;j<n;j++)
			indegree[i] += graph[j][i];
		//cout<<i<<":"<<indegree[i]<<endl;
	}
	for(k=0;k<n;k++)
	{
		for(i=0;i<n && indegree[i]!=0;i++);
		if(i==n) 
			return false;
		indegree[i] = -1;
		path[k] = i;
		for(j=0;j<n;j++)
			indegree[j] -= graph[i][j];
	}
	return true;
}

bool Floyd()
{
	int i,j,k;
	bool map[30][30];
	
    for(i=0;i<n;i++)
		for(j=0;j<n;j++)
			map[i][j]=graph[i][j];
	/*cout<<"1:"<<endl;
    for(i=0;i<n;i++)
	{
		for(j=0;j<n;j++)
			cout<<graph[i][j]<<" ";
		cout<<endl;
	}*/
	for(k=0;k<n;k++)
		for(i=0;i<n;i++)
			for(j=0;j<n;j++)
				if(i!=k && k!=j && map[i][k] && map[k][j])
					map[i][j] = 1;
	/*cout<<"2:"<<endl;
	for(i=0;i<n;i++)
	{
		for(j=0;j<n;j++)
			cout<<map[i][j]<<" ";
		cout<<endl;
	}*/
	bool vis[30],finish=false;
	memset(vis,false,sizeof(vis));
	int totalIndegree = 0;
	for(i=0;i<n;i++)
	{
		totalIndegree = 0;
		for(j=0;j<n;j++)
			totalIndegree += map[j][i];
		//cout<<":"<<totalIndegree<<endl;
		if(vis[totalIndegree])
			return false;
		vis[totalIndegree] = true;
	}
	return true;
}

int main()
{
	char c1,c2,c3;
	bool finish;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		if(n == 0 && m == 0)
			break;
		finish = false;
		memset(graph,0,sizeof(graph));
		for(int i=1;i<=m;i++)
		{
			cin>>c1>>c2>>c3;
			if(finish) continue;				//if it finished before,just input the data and do nothing
			int u = c1-'A';
			int v = c3-'A';
			if(graph[u][v] == true) continue;   //this relationship have already existed!!!!
			graph[u][v] = true;

			if(TopSort() == 0)					//TopSort firstly,if a loop exist,inconsistency found
			{
				finish = true;					//set the flag finish as true
				printf("Inconsistency found after %d relations.\n",i);
				continue;
			}

			if(Floyd())							//judge..
			{
				finish = true;
				printf("Sorted sequence determined after %d relations: ",i);
				for(int j=0;j<n;j++)
					printf("%c",'A'+path[j]);
				cout<<"."<<endl;
				continue;
			}
		}
		if(!finish)
			printf("Sorted sequence cannot be determined.\n");
	}
}