POJ 1094 Sorting It All Out（拓扑排序 入度性质）

原题

Sorting It All Out

Time Limit: 1000MS Memory Limit: 10000K

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 6
A<B
A<C
B<C
C<D
B<D
A<B
3 2
A<B
B<A
26 1
A<Z
0 0

Sample Output

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

题意

依次给定一组字母的大小关系，并及时判断它们是否能组成唯一的拓扑序列。

涉及知识及算法

题解的强大之处在于灵活利用了入度这个概念，存在解的时候，必须满足只有一个单源点，沿着解这条路走，必然是单源点（入度为零）后面跟着入度为1的点， 然后删除单源点，更新入度，重复，又是仅有一个单源点，且后面跟着入度为1的点，最后仅剩一个单源点。有环的话，随着单源点的删除，最后会出现没有单源点的情况，这个时候就可以判读处理有环。 无序的话，无论什么时候，出现多个单源点都说明无序。

——引自CSDN博主chchlh，附上链接http://blog.csdn.net/chchlh/article/details/41846509，表示感谢。

代码

#include <iostream>#include <cstdio>#include <cstring>using namespace std;int n,m;//最终判断的标记bool Sign;//录入字符串string str;//入度数组int indegree[27];//图int Map[27][27];//解空间(队列)int q[27];//备份入度顶点int temp[27];//拓扑排序int TopoSort(){    //记录解空间中零入度顶点的个数    int Count=0;    //记录任一个零入度顶点的位置    int loc;    //记录当前图中零入度顶点数目    int m;    //暂标记为有序    int flag=1;    for(int i=1;i<=n;i++)    {        temp[i]=indegree[i];    }    for(int i=1;i<=n;i++)    {        m=0;        //查找零入度顶点个数        for(int j=1;j<=n;j++)        {            if(temp[j]==0)            {                m++;                //记录一个零入度顶点位置                loc=j;            }        }        //当前图中零入度顶点数目为零一定说明有环        if(m==0)        {            return 0;        }        //无序，但不一定知道是否有环        if(m>1)        {            flag=-1;        }        //该零入度顶点入队        q[Count++]=loc;        //入度置为-1        temp[loc]=-1;        //删除该点        for(int j=1;j<=n;j++)        {            if(Map[loc][j]==1)            {                temp[j]--;            }        }    }    return flag;}int main(){    //freopen("in.txt","r",stdin);    while(scanf("%d%d",&n,&m)&&(n||m))    {        memset(Map,0,sizeof(Map));        memset(indegree,0,sizeof(indegree));        Sign=0;        for(int i=1;i<=m;i++)        {            cin>>str;            //一旦确定结果，就对后续的输入不再操作            if(Sign)            {                continue;            }            int u=str[0]-'A'+1;            int v=str[2]-'A'+1;            Map[u][v]=1;            //入度加一            indegree[v]++;            int s=TopoSort();            //有环            if(s==0)            {                printf("Inconsistency found after %d relations.\n",i);                Sign=1;            }            //有序            else if(s==1)            {                printf("Sorted sequence determined after %d relations: ",i);                for (int j=0; j<n; j++)                {                    putchar(q[j]+'A'-1); //输出字符 putchar(ASCII)                }                printf(".\n");                Sign=1;            }        }        //无法得出结果        if(!Sign)        {             printf("Sorted sequence cannot be determined.\n");        }    }    return 0;} /**************************************************************    Language: C++    Result: Accepted    Time:0 ms    Memory:1508 kb****************************************************************/

代码引自新浪博客博主康文骐，附上链接http://blog.sina.com.cn/s/blog_676070110100kii1.html，表示感谢。