POJ 1094 拓扑排序

Sorting It All Out

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 24712		Accepted: 8565

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.

题意：根据题目所给的字母关系，确定拓扑排序情况。

解题思路：由于此题要输出在多少条关系下发生冲突，或者排序关系确定，或者所有的关系，或者仍然有不确定的关系，所以每加入一条边，拓扑排序一次，知道排序完成，或者发现冲突，或者加完m条边。

假如存在不确定关系，也存在冲突，那么冲突的优先级要高。

代码：

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string>
#include<string.h>
#include<queue>
using namespace std;
int in[50],head[50],tol,num[50],cnt,m,n;
struct node
{
        int next,to;
}edge[10000];
struct Node
{
        int st,ed;
}pp[400];
void add(int u,int v)
{
        edge[tol].to=v;
        edge[tol].next=head[u];
        head[u]=tol++;
}
int topo()
{
        int i,j,k;
        int ind[100];
        cnt=0;
        int flag=1;
        queue<int> q;
        for(i=0;i<n;i++)
        {
                ind[i]=in[i];
                if(ind[i]==0)q.push(i);
        }
        while(!q.empty())
        {
                if(q.size()>1)flag=0;
                int u=q.front();q.pop();
                num[cnt++]=u;
                ind[u]--;
                for(i=head[u];i!=-1;i=edge[i].next)
                {
                        int v=edge[i].to;
                        if(--ind[v]==0)q.push(v);
                }
        }
        //cout<<cnt<<endl;
        if(cnt<n)return -1;
        if(flag==0)return 0;
        return 1;
}
void solve()
{
        int i,j,k,p,q;
        memset(in,0,sizeof(in));
        memset(head,-1,sizeof(head));tol=0;
        for(p=1;p<=m;p++)
        {
               add(pp[p].st,pp[p].ed);
               in[pp[p].ed]++;
               q=topo();
               if(q==-1)
               {
                       printf("Inconsistency found after %d relations.\n",p);
                       return;
               }
               if(q==1)
               {
                       printf("Sorted sequence determined after %d relations: ",p);
                       for(i=0;i<cnt;i++)printf("%c",num[i]+'A');
                       puts(".");
                       return;
               }
        }
        puts("Sorted sequence cannot be determined.");
}
int main()
{
        int i,j,k;
        char str[10];
        //freopen("3.in","r",stdin);
       // freopen("data.out","w",stdout);
        while(~scanf("%d%d",&n,&m))
        {
                if(n==0&&m==0)break;
                for(i=1;i<=m;i++)
                {
                      scanf("%s",str);
                      int t1=str[0]-'A';
                      int t2=str[2]-'A';
                      if(str[1]=='>')pp[i].st=t2,pp[i].ed=t1;
                      else pp[i].st=t1,pp[i].ed=t2;
                }
                solve();
             //  for(i=1;i<=m;i++)cout<<pp[i].st<<" "<<pp[i].ed<<endl;
        }
        return 0;
}