POJ Sorting It All Out 1094(拓扑排序)

                                                                                        Sorting It All Out

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 32801 Accepted: 11394

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.

题意：１给出m条关系，如果可以确定一个唯一的关系，则把这个关系输出来

　　　２如果在i条关系存在环，则输出在i条关系存在环

　　　３其他情况输出不确定

每给出一条关系都需要判断一下，如果有１或２的情况，则只录入，不判断

拓扑排序，用一个队列，如果度等于０，入队，然后找以这个点为前缀的点，改变度，直到队列为空

#include <stdio.h>#include <string.h>#include <stdlib.h>#include <queue>#include <algorithm>using namespace std;int v[50],tep[50],du[50],mp[50][50],k[50];int n;void init(){    for(int i=0;i<n;i++)    {        du[i]=0;    }    memset(mp,0,sizeof(mp));}int topsort(){    memset(v,0,sizeof(v));    queue<int >q;    while(!q.empty())q.pop();    int i,j;    for(i=0;i<n;i++)    {        if(du[i]==0)        {            v[i]=1;            q.push(i);        }    }    int ff=0,pos=0;    while(!q.empty())    {        if(q.size()>1)ff=1;        int t=q.front();        q.pop();        k[pos++]=t;        for(i=0;i<n;i++)        {            if(mp[t][i])                du[i]--;        }        for(i=0;i<n;i++)        {            if(!v[i] && du[i]==0)            {                v[i]=1;                q.push(i);            }        }    }    if(pos<n) return 1;    if(ff) return 2;    return 3;}int main(){    int m,t,i,j;    while(~scanf("%d%d",&n,&m))    {        if(n==0&&m==0)        {            break;        }        init();        char s[10];        int flag=2,ff=0,stop;        for(j=1;j<=m;j++)        {            scanf("%s",s);            if(ff)continue;            int a=s[0]-'A';            int b=s[2]-'A';            if(!mp[a][b])            {                mp[a][b]=1;                du[b]++;            }            for(i=0;i<n;i++)                tep[i]=du[i];            flag=topsort();            for(i=0;i<n;i++)                du[i]=tep[i];            if(flag!=2)            {                ff=1;                stop=j;            }        }        if(flag==3)        {            printf("Sorted sequence determined after %d relations: ",stop);            for(i=0;i<n;i++)            {                printf("%c",k[i]+'A');            }            printf(".\n");        }        else if(flag==1)        {            printf("Inconsistency found after %d relations.\n",stop);        }        else        {            printf("Sorted sequence cannot be determined.\n");        }    }}