poj1094 topsort

Sorting It All Out

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 32275		Accepted: 11210

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：给你字母表的前N个字母以及M个他们之间的关系，问最终是全序，偏序，还是出现矛盾。

思路分析：有向图，很明显拓扑排序，这是我做的第二道拓扑排序的题目，在学习图论之前一直以为拓扑

排序就是单纯的一种排序方式，现在才知道它是图论的一部分，有向图，因为本题不仅要求输出结果，还要判断

在输入第几个顺序时得出的结果，因此每输入一组数据就要进行一拓扑排序，拓扑排序本质上一种偏序排序，

当然在一定情况下会出现全序的特殊情况，j!=n，就说明有向图的构建过程中出现了环，有可能直接是反向环，

也有可能是绕一圈后回来，如果队列中元素>1，说明队列中存在两个以上优先级相同的元素，即有向图是偏序的。

不说了，上代码，本题对于拓扑排序的理解还是十分有用的。

代码：

#include <iostream>
#include <algorithm>
#include <stack>
#include <queue>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn=27;
int indegree[maxn],ans[maxn],graph[maxn][maxn];
int n,m;
int topsort()
{
    int flag=0;
    int in[maxn];
    int j=0;
    memcpy(in,indegree,sizeof(indegree));//一定要重新定义一个数组，防止对主函数中indegree数组产生影响
    queue<int> q;
    for(int i=0;i<n;i++)
    {
        if(in[i]==0) q.push(i);//将入度为0的点放入队列
    }
    while(!q.empty())
    {
        if(q.size()>1)  flag=1;
        int t=q.front();
        q.pop();//记得弹出
        ans[j++]=t;
        for(int i=0;i<n;i++)
        {
            if(graph[t][i])
            {
                in[i]--;
                if(in[i]==0) q.push(i);
            }
        }
    }
    if(j!=n)//不能拓扑排序 ,存在环
        return 1;
    else if(flag==1) return 2;
    else return 0;
}
int main()
{
    char s[5];
    while(scanf("%d%d",&n,&m)&&(n||m))
    {
        int incon=0,detmin=0;
        memset(graph,0,sizeof(graph));
        memset(indegree,0,sizeof(indegree));
        for(int i=1;i<=m;i++)
        {
            scanf("%s",s);
            if(detmin||incon) continue;
            int a=s[0]-'A';
            int b=s[2]-'A';
            if(graph[a][b]==0)
            {
                graph[a][b]=1;
                indegree[b]++;
            }
            int res=topsort();
            if(res==0)
            {
                detmin=1;
                printf("Sorted sequence determined after %d relations: ",i);
                for(int i=0;i<n;i++)
                printf("%c",ans[i]+'A');
                printf(". ");
            }
            if(res==1)
            {
                incon=1;
                printf("Inconsistency found after %d relations. ",i);
            }
        }
        if(!incon&&!detmin) printf("Sorted sequence cannot be determined. ");
    }

}