poj 1094 Sorting It All Out 初级-》图算法-》拓扑排序

题意：一直到给定的第几条指令可以唯一确定一个拓扑排序的结果。 结果有三种：①、可以唯一确定；②、有环也就是矛盾；③、不能唯一确定也就是不能确定。 此题需要我们输入一条指令，执行一次拓扑排序。执行完一次拓扑排序后记得把入度再加回去。

 

[align=center]Sorting It All Out[/align]

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 22173		Accepted: 7647

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.

 

 

 

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<stack>
#include<vector>
using namespace std;
vector <int> G[100];
stack<int>st1,st2;
int main()
{

int n,m,i,j,k,len;
int c1,c2,f1,f2,l1,l2,l3;
int in[28];
char ch[5],str[28];

while(scanf("%d%d",&n,&m),n+m)
{
memset(in,0,sizeof(in));
for(i=0;i<=n;i++) // 初始化
G[i].clear();
f1=0;
for(i=1;i<=m;i++)
{
scanf("%s",ch);
G[ch[0]-'A'].push_back(ch[2]-'A');
in[ch[2]-'A']++;

if(f1==0)
{
c1=0; f2=0; k=0;
for(j=0;j<n;j++)
if(in[j]==0)
{
st1.push(j);
st2.push(j);
}
while(st1.size())
{
if(st1.size()>=2) f2=1;
l1=st1.top();
str[k++]=l1+'A';
c1++; //记录一次拓扑排序后入度为零的点的个数
st1.pop();
len=G[l1].size();
for(j=0;j<len;j++)
{
l2=G[l1][j];
in[l2]--;
if(in[l2]==0)
{
st1.push(l2);
st2.push(l2);
}
}
}
if(c1!=n)
{
f1=1; c2=i;
}
else
{
if(c1==n&&f2==0)
{
f1=3; c2=i;
}
else if(i==m&&f2==1)
{
f1=2;
}
}
while(st2.size())
{
l2=st2.top();
st2.pop();
len=G[l2].size();
for(j=0;j<len;j++)
{
l3=G[l2][j];
in[l3]++;
}
}
}
}
if(f1==1)
{
cout<<"Inconsistency found after ";
cout<<c2<<" relations."<<endl;
}
if(f1==2)
{
cout<<"Sorted sequence cannot be ";
cout<<"determined."<<endl;
}
if(f1==3)
{
str
='\0';
cout<<"Sorted sequence determined after";
cout<<" "<<c2<<" relations: ";
cout<<str<<"."<<endl;
}
}
return 0;
}