字符拓扑 POJ 1094

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.

#include <iostream>#include <cstring>#include <cstdio>using namespace std;int indegree[27];     //入度int map[27][27];     //图int queue[27];       // 每次找到入度为0的点后，存储要输出的顺序int TopoSort(int n)     //拓扑排序{    int count=0;     //记录解空间中零入度顶点的个数    int temp[27];     //对入度顶点备份    int pos;      //记录一个零入度顶点的数目    int i,j;    int m;       //零入度顶点的个数    int flag=1;      //flag=1：有序  flag=-1：不能确定    for (i=1; i<=n; i++)    {        temp[i] = indegree[i];  //备份    }    for (i=1; i<=n; i++)  //遍历n遍,必须把图全部遍历完（根据当前输入的，尽管map全部初始化为0了，只要根据当前输入的所有条件判断）    {        m = 0;        for (j=1; j<=n; j++)  //查找零入度顶点的个数        {            if (temp[j] == 0)            {                m ++;                pos = j;   //记录一个零入度顶点的位置            }        }        if (m == 0)     //零入度顶点的个数==0：有环        {            return 0;        }  if (m > 1)     //零入度顶点的个数>1,说明还有其他两点之间没确定关系，判断是无序（只有1到n个点确定关系，且没有环，根据拓扑排序才能确定有序）  {            flag=-1;    //当知道无序时，并不一定知道该图是否有环，因此要继续遍历，可别急着退出...        }        queue[count++] = pos;   //零入度顶点入队        temp[pos] = -1;    //将零入度顶点的入度置为-1        for (j=1; j<=n; j++)  //删除以pos为起点的边        {            if (map[pos][j] == 1)            {                temp[j] --;   //相应定点的入度减1            }        }    }    return flag;}int main(){    int n,m;    int sign;      //当sign=1时，程序已经得出结果，不需再考虑后面的输入    string str;    while (scanf("%d%d",&n,&m) && n!=0 && m!=0)    {        memset(map,0,sizeof(map));        memset(indegree,0,sizeof(indegree));        sign=0;        for (int i=1; i<=m; i++)        {            cin>>str;            if (sign)            {                continue;   //一旦得出结果，对后续的输入置之不理！            }            int u = str[0]-'A'+1;            int v = str[2]-'A'+1;            map[u][v] = 1;            indegree[v] ++;            int s = TopoSort(n);            if (s == 0)    //有环            {                printf("Inconsistency found after %d relations.\n",i);                sign=1;            }            if (s == 1)    //有序            {                printf("Sorted sequence determined after %d relations: ",i);                for (int j=0; j<n; j++)                {                    printf("%c",queue[j]+'A'-1); //根据进队顺序输出字符，整型转字符型                }                printf(".\n");//每句话的结尾的点号，记得输出！                sign=1;            }        }        if(!sign)     //无法得出结果        {            printf("Sorted sequence cannot be determined.\n");        }    }    return 0;}