Sorting It All Out（拓扑排序）

题目：

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：输入n,m，n和m为0的时候输入结束，n代表n个字母，接下来输入m组数据，然后让你根据这m组数据判断是否能将这m个字母排序好，要注意在第i组数据能排序好的时候就输出结果，接下来的数据值输入，不进行判断；

分析：这是我第一次做有于拓扑排序有关的题目，根据题意我们需要分为三步：①判断是否形成循环 ②判断是否有序 ③无法判断（题目输入的数据不足以判断）

AC代码：

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
#define N 30
char Map[N][N];
int  indegree[N],p[N];
int c;
int toposort(int n)
{
     c=0;
     int sign=1;
    int temp[N],m,now;
    for (int i=1;i<=n;i++)
        temp[i]=indegree[i];
    for (int i=1;i<=n;i++)
    {
        m=0;
        for (int j=1;j<=n;j++)
            if (temp[j]==0)
                m++,now=j;
        if (m==0)
            return 0;   //成环
        if (m>1)
            sign=-1;   //无法判断
        p[c++]=now;
        temp[now]=-1;
        for (int j=1;j<=n;j++)
            if (Map[now][j]==1)
               temp[j]--;
    }
    return sign;  //有序
}
int main()
{
    int n,m;
    char w[5];
    int x,y;
    while (scanf("%d%d",&n,&m)&&(n!=0||m!=0))
    {
        int flag=0;
        memset(Map,0,sizeof(Map));
        memset(indegree,0,sizeof(indegree));
        for (int i=1;i<=m;i++)
        {

            scanf("%s",w);
            if (flag)
                continue;
            x=w[0]-'A'+1;
            y=w[2]-'A'+1;
            Map[x][y]=1;     //x->y
            indegree[y]++;   //后者入度+1
            int t=toposort(n);
            if (t==0)
                printf("Inconsistency found after %d relations.
",i),flag=1;
            else if (t==1)
            {
                printf("Sorted sequence determined after %d relations: ",i);
                for(int j=0;j<c;j++)
                    printf("%c",p[j]+'A'-1);
                    printf(".
");
                    flag=1;
            }
        }
        if (!flag)
        printf("Sorted sequence cannot be determined.
");
    }
    return 0;
}