POJ 1094 Sorting It All Out

Sorting It All Out

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 22346		Accepted: 7710

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

题意：给你一堆类似于A<B,A<C之类的关系，最后让你归结出前N个字母的升序序列是什么情况。有三种情况：1.确定了只有一种升序关系；2.关系之间发生冲突；3.有多重不确定的关系。

解题思路：从题意来看，最后需要输出关系序列，那么关系中只有<的关系，可以用拓扑排序来求他们的关系序列，对于A<B,从A到B连接一条又向边，每次连接都判断一下当前的关系状态。

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
const int maxn=30;
const int maxm=100000;
const int oo=1000000000;
int head[maxn],deg[maxn],que[maxn],tmpdeg[maxn];
int next[maxm],ver[maxm];
int edge,n,m,sign,l,r;
bool vis[maxn];
void init()
{
    for(int i=0; i<n; i++)
        head[i]=-1,deg[i]=0;
    edge=0;
}
inline void add(int u,int v)
{
    ver[edge]=v,next[edge]=head[u],head[u]=edge++;
    deg[v]++;
}
int TopSort()
{
    l=r=0;
    int f=1;
    memset(que,-1,sizeof(que));
    memset(vis,0,sizeof(vis));
    for(int i=0; i<n; i++)
        tmpdeg[i]=deg[i];
    for(int i=0; i<n; i++)
        if(tmpdeg[i]==0)
        {
            vis[i]=1;
            que[r++]=i;
        }
    if(r>1) f=2;
    while(l<r)
    {
        int top=que[l++];
        for(int i=head[top]; i!=-1; i=next[i])
            tmpdeg[ver[i]]--;
        for(int i=0; i<n; i++)
            if(tmpdeg[i]==0&&!vis[i])
            {
                vis[i]=1;
                que[r++]=i;
            }
        if(r-l>1) f=2;
    }
    if(r<n) return 0;
    return f;
}
int main()
{
    char op[5];
    int ans=0,tmp;
    while(cin>>n>>m)
    {
        if(n==0&&m==0) break;
        init();
        int flag=0;
        ans=0;
        for(int i=0; i<m; i++)
        {
            cin>>op;
            if(flag>0) continue;
            add((int)(op[0]-'A'),(int)(op[2]-'A'));
            if((tmp=TopSort())==0)
            {
                ans=-1;
                flag=i+1;
            }
            else if(tmp==1) flag=i+1;
        }
        if(flag)
        {
            if(ans!=-1)
            {
                cout<<"Sorted sequence determined after "<<flag<<" relations: ";
                for(int i=0; i<n; i++)
                    cout<<(char)(que[i]+'A');
                cout<<"."<<endl;
            }
            else cout<<"Inconsistency found after "<<flag<<" relations."<<endl;
        }
        else cout<<"Sorted sequence cannot be determined."<<endl;
    }
    return 0;
}