POJ 1094：Sorting It All Out（拓扑排序+传递闭包）

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

对前n个字母，输入m组确定他们之间的大小关系，从而判断有没有唯一的序列。

第一种输出，在前t(t<=m)已能确定一个序列

第二种:不能确定唯一的序列

第三种：不存在拓扑排序

思路：拓扑排序+floyd。先用拓扑排序判断在当前情况下，有没有环的存在，如果有环的存在则证明m组关系是有矛盾的，

即可输出3。若没有环，那没接下来就是判断能否构成唯一的有序序列，用floyd传递闭包，如果能构成唯一的有序序列，

则顶点1~n，必分布着0~n-1个入度，才能保证0个入度的顶点排在第1位，1个入度的顶点排在第2位（因为只有1个顶点比它大），

以此类推。有一点要注意下，就是要每输入一次判断一下，如果有输出了，以后就只输入就好，不用第二次输出。

源代码：

#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");
}
}

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