POJ1094 Sorting It All Out（拓扑排序）每输入条关系判断一次

Sorting It All Out

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 29182 Accepted: 10109

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 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.

Source

East Central North America 2001

#include<stdio.h>const int N = 30;int mapt[N][N],in[N],n,m;int tope(int tm){    int path[N],k=0,l=0,uncertain=0;    for(int i=0;i<n;i++)    if(in[i]==0)    path[k++]=i,in[i]=-1;    while(l<k)    {        int s=path[l++];        if(k-l!=0)            uncertain=1;        for(int j=0;j<n;j++)        if(in[j]>0&&mapt[s][j])        {            in[j]--;            if(in[j]==0)                path[k++]=j,in[j]=-1;        }    }    int flag=0;    if(k!=n)//说明有环，矛盾        printf("Inconsistency found after %d relations.\n",tm),flag=1;    else if(uncertain==0)    {        printf("Sorted sequence determined after %d relations: ",tm);        for(int i=0;i<k;i++)            printf("%c",path[i]+'A');        printf(".\n");        flag=1;    }    else if(tm==m)//最后一次加入一个关系判断，可能的输出        printf("Sorted sequence cannot be determined.\n"),flag=1;    return flag;}int main(){    int in_t[N];    char st[10];    while(scanf("%d%d",&n,&m)>0&&n+m!=0)    {        for(int i=0;i<n;i++)        {            in_t[i]=0;            for(int j=0;j<n;j++)                mapt[i][j]=0;        }        int flag=0;        for(int i=1;i<=m;i++)        {            scanf("%s",st);            if(flag)                continue;            int a=st[0]-'A';            int b=st[2]-'A';            if(mapt[a][b]==0)            {                mapt[a][b]=1; in_t[b]++;                for(int j=0;j<n;j++)                    in[j]=in_t[j];                flag=tope(i);            }        }    }}