A

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 6A<BA<CB<CC<DB<DA<B3 2A<BB<A26 1A<Z0 0

Sample Output

Sorted sequence determined after 4 relations: ABCD.Inconsistency found after 2 relations.Sorted sequence cannot be determined.  

分析：
这题典型的拓扑排序，但是有点变化。
题目样例的三种输出分别是：
1. 在第x个关系中可以唯一的确定排序，并输出。
2. 在第x个关系中发现了有回环（Inconsisitency矛盾）
3.全部关系都没有发现上面两种情况，输出第3种.
贪心的做法：
1. 找到所有入度为0的点， 加入队列Q
2.取出队列Q的一个点，把以这个点为起点，所有它的终点的入度都减1. 如果这个过程中发现经过减1后入度变为0的，把这个点加入队列Q。
3.重复步骤2，直到Q为空。
这个过程中，如果同时有多个点的入度为0，说明不能唯一确定关系。
如果结束之后，所得到的经过排序的点少于点的总数，那么说明有回环

刚开始时没看懂样例输入输出，后来懂了又在确定唯一关系时出问题（我是用输入的不同字母的个数是否达到n来判断是否达到唯一确定的关系）；       
#include<cstdio>#include<cstring>#include<vector>#include<queue>using namespace std;const int N = 105;int n,m,in[N],temp[N],Sort[N],t,pos, num,fa[30][30];char X, O, Y;vector<int>G[N];void init(){    memset(in, 0, sizeof(in));    for(int i=0; i<=n; ++i){        G[i].clear();    }}int topoSort(){    queue<int>q;    for(int i=0; i<n; ++i)if(in[i]==0){            q.push(i);    }    pos=0;    bool unSure=false;    while(!q.empty()){        if(q.size()>1) unSure=true;        int t=q.front();        q.pop();        Sort[pos++]=t;        for(int i=0; i<G[t].size(); ++i){            if(--in[G[t][i]]==0)                q.push(G[t][i]);        }    }    if(pos<n) return 1;    if(unSure)  return 2;    return 3;}int main(){    int x,y,i,flag,ok,stop;    while(~scanf("%d%d%*c",&n,&m)){        if(!n||!m)break;memset(fa,0,sizeof(fa));        init();        flag=2;        ok=false;        for(i=1; i<=m; ++i){            scanf("%c%c%c%*c", &X,&O,&Y);            if(ok||fa[X-'A'][Y-'A']) continue; // 如果已经判断了有回环或者可唯一排序，不处理但是要继续读            x=X-'A', y=Y-'A';fa[x][y]=1;            G[y].push_back(x);                    ++in[x];            // 拷贝一个副本，等下用来还原in数组            memcpy(temp, in, sizeof(in));            flag=topoSort();            memcpy(in, temp, sizeof(temp));            if(flag!=2){                stop=i;                ok=true;            }        }        if(flag==3){            printf("Sorted sequence determined after %d relations: ", stop);            for(int i=pos-1; i>=0; --i)                printf("%c",Sort[i]+'A');            printf(".\n");        }        else if(flag==1){            printf("Inconsistency found after %d relations.\n",stop);        }        else{            printf("Sorted sequence cannot be determined.\n");        }    }    return 0;}