链接：http://poj.org/problem?id=1094

题意&思路：直接拓扑排序。多解输出一串英文，有环输出一段英文，唯一解输出一段英文及排序结果。

细节：题目描述不是很清楚，如果不看discuss我肯定要WA出翔。

discuss里总结了两点关键的：

1. 输入一条边时如果此时拓扑有解就输出这个解，即使后面的边成有向环也不管了，所以每次输入的时候都得进行拓扑排序。

2. 判断存在有向环应先于判断多解。

这道题主要是题目坑爹。

#include<cstring>
#include<string>
#include<fstream>
#include<iostream>
#include<iomanip>
#include<cstdio>
#include<cctype>
#include<algorithm>
#include<queue>
#include<map>
#include<set>
#include<vector>
#include<stack>
#include<ctime>
#include<cstdlib>
#include<functional>
#include<cmath>
using namespace std;
#define PI acos(-1.0)
#define MAXN 50100
#define eps 1e-7
#define INF 0x7FFFFFFF
#define LLINF 0x7FFFFFFFFFFFFFFF
#define seed 131
#define MOD 1000000007
#define ll long long
#define ull unsigned ll
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1

struct node{
    int u,v,next;
}edge[MAXN];
int head[30],in[30],ans[30],IN[30];
int n,m,cnt,sum;
void add_edge(int a,int b){
    edge[cnt].u = a;
    edge[cnt].v = b;
    edge[cnt].next = head[a];
    head[a] = cnt++;
}
int toposort(){
    int i;
    for(i=0;i<n;i++){
        in[i] = IN[i];
    }
    queue<int>q;
    for(i=0;i<n;i++){
        if(!in[i]) q.push(i);
    }
    sum = 0;
    int flag = 0;
    if(q.size()>1)  flag = 1;
    while(!q.empty()){
        int u = q.front();
        q.pop();
        ans[sum] = u;
        sum++;
        for(i=head[u];i!=-1;i=edge[i].next){
            int next = edge[i].v;
            in[next]--;
            if(in[next] == 0)   q.push(next);
        }
        if(q.size()>1)  flag = 1;
    }
    if(sum!=n)  return 2;
    else if(flag)   return 1;
    return 3;
}
int main(){
    int i,j;
    char str[5];
    while(scanf("%d%d",&n,&m),n||m){
        memset(IN,0,sizeof(in));
        memset(head,-1,sizeof(head));
        int flag = 0;
        cnt = 0;
        int temp;
        int p;
        for(i=0;i<m;i++){
            scanf("%s",str);
//            cout<<str<<endl;
            if(flag)    continue;
            IN[str[2]-'A']++;
            add_edge(str[0]-'A',str[2]-'A');
            temp = toposort();
            if(temp==3||temp==2){
                flag = 1;
                p = i + 1;
            }
        }
        if(flag){
            if(temp==3){
                printf("Sorted sequence determined after %d relations: ",p);
                    for(i=0;i<n;i++){
                        printf("%c",ans[i]+'A');
                    }
                printf(".\n");
            }
            else if(temp==2){
                printf("Inconsistency found after %d relations.\n",p);
            }
        }
        else    printf("Sorted sequence cannot be determined.\n");
    }
    return 0;
}

POJ--1094--Sorting It All Out【拓扑排序】