POJ - 1094 Sorting It All Out （拓扑排序）

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：看第几行边就能看出有唯一的顺序或者   成环   ，如果所有边加完后都没有唯一顺序也没成环，那么输出 

Sorted sequence cannot be determined.

解：每加新的边，都进行1次拓扑排序 

#include<stdio.h>
#include<string>
#include<string.h>
#include<iostream>
#include<math.h>
#include<queue>
#include<vector>
#include<algorithm>
#include<map>d
#define inf 0x3f3f3f3f
#define ll long long
#define maxx 5000005
using namespace std;
int n,m,tot,flag,cnt,flag2;
struct node
{
    int en,next;
}e[10000];
int p[10000];
int a[10000];
int b[10000];
int vis[10000];
int ans[10000];
void add(int u,int v)
{
    e[tot].en=v;
    e[tot].next=p[u];
    p[u]=tot++;
}
void aa(int u)
{
    int kk=0;
    int pp=0;
    queue<int>q;
    for(int i=0;i<n;i++)
        if(vis[i]&&!b[i]) q.push(i);
    while(!q.empty())
    {
        int x=q.front();
        q.pop();
        if(!q.empty()) kk=1;
        ans[pp++]=x;
        for(int i=p[x];i+1;i=e[i].next)
        {
            int y=e[i].en;
            b[y]--;
            if(!b[y]) q.push(y);
        }
    }
    if(pp!=cnt)
        flag=u;
    if(!kk&&pp==n)
        flag2=u;
}
int main()
{
    while(~scanf("%d%d",&n,&m)&&n+m)
    {
        char s[10];
       memset(p,-1,sizeof(p));
       memset(a,0,sizeof(a));
       memset(vis,0,sizeof(vis));
        tot=0;
        cnt=0;
        flag=0;
        flag2=0;
        for(int i=1;i<=m;i++)
        {
           scanf("%s",s);
           if(flag) continue;
           if(flag2) continue;
           int u=s[0]-'A';
           int v=s[2]-'A';
           if(!vis[u])
           {
               vis[u]=1;
               cnt++;
           }
           if(!vis[v])
           {
               vis[v]=1;
               cnt++;
           }
           add(u,v);
           a[v]++;
           memcpy(b,a,sizeof(a));
           aa(i);
        }
        if(flag)
        {
            printf("Inconsistency found after %d relations.\n",flag);
            continue;
        }
        if(flag2)
        {
            printf("Sorted sequence determined after %d relations: ",flag2);
            for(int i=0;i<n;i++)
                printf("%c",ans[i]+'A');
            printf(".\n");
            continue;
        }
        printf("Sorted sequence cannot be determined.\n");


    }
}