poj1094

Sorting It All Out

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 32966		Accepted: 11458

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.

题目大意：给你n个大写字母，和m组字母相互之间的大小关系，让你判断在哪一次能够将所有字母
排好序，又或者在哪一次给出的关系与前面矛盾，即构成环了，否则就不能判断

思路：利用拓扑排序，如果能够排序，那么入度为0的点只可能有一个，否则就不知道选择哪一个在前
如果所有的点不能全部进行出栈排序，说明一定形成环了
注意：在拓扑排序时一定要注意判断顺序，就因为这个WA了好多次

代码如下：

#include <iostream>
#include<cstdio>
#include<stack>
#include<cstring>
using namespace std;
int n,m;
char str[4];
int degree[27],A[27][27];//入度，以及邻接矩阵
char letter[27];//存储排序好的字母
int topSort()
{
    //如果能够排序，那么肯定入度为0的点只有一个,有多个则不能确定选择哪一个
    stack<int> st;
    int temp[27],flag=0;
    memcpy(temp,degree,sizeof(degree));
    for(int i=1;i<=n;i++)
        if(temp[i]==0)
            st.push(i);//找到入度为0的点
    int index=0,cur;
    while(!st.empty())
    {
        if(st.size()>1)//入度为0点有多个，不能唯一排序
            flag=1;
        cur = st.top();
        st.pop();
        letter[index++]=cur+'A'-1;
        for(int i=1;i<=n;i++)
            if(A[cur][i]==1)
            {
                temp[i]--;
                if(temp[i]==0)
                    st.push(i);//找到入度为0的点
            }
    }
    if(index<n)
        return 0;//有环
    if(flag)
        return -1;
    return 1;//排序成功

}
int main()
{
    freopen("in1.txt","r",stdin);
    while(scanf("%d%d",&n,&m)!=EOF&&n)
    {
        bool sucess=false;
        int flag=-1,time=0;
        memset(degree,0,sizeof(degree));
        memset(A,0,sizeof(A));
        memset(letter,0,sizeof(letter));
        if(m==0&&n==1)
        {
            printf("Sorted sequence determined after 0 relations: A.\n");
            continue;
        }
        for(int i=1;i<=m;i++)
        {
            scanf("%s",str);
            if(sucess)
                continue;
            int x = str[0]-'A'+1;
            int y = str[2]-'A'+1;
            degree[y]++;//入度加1
            A[x][y]=1;//表示x邻接y
            int flag1 = topSort();
            if(flag1==0)
            {
                flag=0;time=i;
                sucess=true;
            }
            if(flag1==1)
            {
                flag=1;time=i;
                sucess=true;
            }

        }
        if(flag==-1)
            printf("Sorted sequence cannot be determined.\n");
        else if(flag==0)
            printf("Inconsistency found after %d relations.\n",time);
        else
        {
             printf("Sorted sequence determined after %d relations: ",time);
             for(int j=0;j<n;j++)
                printf("%c",letter[j]);
             printf(".\n");
        }
    }
    return 0;
}