POJ1094 Sorting It All Out —— 拓扑排序

题目链接：http://poj.org/problem?id=1094

Sorting It All Out

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 35468		Accepted: 12458

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.

题解：

1.由于要输出当关系确定时是第几步，所以每一步都需要进行拓扑排序。（假如读完所有数据再进行拓扑排序，则无法确定关系确定时是第几步）。

2.在进行拓扑排序时（flag为状态类型，1表示冲突； 2表示不能确定； 3表示关系确定。 flag初始化为3）：

(1) 如果入度为0的点的个数为0，则成环，可直接得出结果：冲突。直接return 1。

(2) 如果入度为0的点的个数为1，继续。

(3) 如果入度为0的点的个数大于1，则表明最小的那个不能确定。这时把flag改为2，然后继续。（为什么还要继续，不是可以直接return 2，表明关系还不能确定了吗？ 虽然如此，但是继续拓扑排序下去，可能会发现环，即冲突。）

代码如下：

#include<stdio.h>//poj 1094
#include<string.h>
#include<stdlib.h>

int deg[27], sub_deg[27],edge[27][27],n,m;
char a,b,c,ans[27];

int TopoSort()
{
    int flag = 3;
    memcpy(sub_deg,deg,sizeof(deg));
    for(int i = 0; i<n; i++)
    {
        int k, cnt = 0;
        for(int j = 0; j<n; j++)
            if(sub_deg[j]==0) cnt++, k = j;

        if(cnt==0) return 1;    //成环
        if(cnt>1) flag = 2;     //最小的不能确定。但不能直接退出，因为接下来可能会出现环。

        sub_deg[k] = -1;
        ans[i] = k+65;
        for(int j = 0; j<n; j++)
            if(edge[k][j]) sub_deg[j]--;
    }
    return flag;
}

int main()
{
    int flag,step;
    while(scanf("%d%d",&n,&m) && (n||m))
    {
        int finished = 0, flag, step;
        memset(deg,0,sizeof(deg));
        memset(edge,0,sizeof(edge));
        for(int i = 1; i<=m; i++)
        {
            getchar();
            scanf("%c%c%c",&a,&c,&b);
            if(finished) continue;
            deg[b-65]++;
            edge[a-65][b-65] = 1;

            ans[n] = 0;
            flag = TopoSort();
            if(flag==1 || flag==3) { step = i; finished = 1; }
        }

        if(flag==1)
            printf("Inconsistency found after %d relations.\n",step);
        else if(flag==2)
            printf("Sorted sequence cannot be determined.\n");
        else
            printf("Sorted sequence determined after %d relations: %s.\n",step,ans);
    }
    return 0;
}