[ACM] POJ 1094 Sorting It All Out (拓扑排序)

Sorting It All Out

Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 26801 Accepted: 9248

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.

Source

East Central North America 2001

解题思路:

拓扑排序的应用。参考http://www.cnblogs.com/pushing-my-way/archive/2012/08/23/2652033.html做的。

本题需要注意的问题很多,有点“坑”。下面是从上面博文中转的,()里面的内容是我自己加的。

题意:给你一些大写字母间的偏序关系,然后让你判断能否唯一确定它们之间的关系,或者所给关系是矛盾的,或者到最后也不能确定它们之间的关系。

分析:

用拓扑排序:

1.拓扑排序可以用栈来实现,每次入栈的是入度为0的节点(也可以用队列,或者不使用队列和栈,循环n次,找入度为0的点)。

1.拓扑排序的结果一般分为三种情况:1、可以判断(拓扑排序有唯一的结果) 2、有环出现了矛盾(出现了没有入度为0的节点) 3、条件不足,不能判断.

2.这道题不仅需要判断这三种情况,而且还要判断在处理第几个关系时出现前两种情况,对于本道题来说三种情况是有优先级的。前两种情况是平等的谁先出现先输出谁的相应结果,对于第三种情况是在前两种情况下都没有的前提下输出相应结果的.

网上对于这道题的错误提示(需要注意):

1.本题顺序:

a.先判有没有环,有环就直接输出不能确定;

b.如果没有环,那么就看会不会有多种情况,如果有多种情况就再读下一行;如果全部行读完还是有多种情况,就是确定不了;

c.如果最后没有环,也不存在多种情况(即每次取出来入度为零的点只有一个),那么才是答案;

2.有答案就先出答案,不管后面的会不会矛盾什么的;

3.如果在没有读完所有输入就能出答案,一定要把剩下的行都读完。

代码:

#include <iostream>
#include <algorithm>
#include <string.h>
#include <stack>
#include <queue>
using namespace std;
int indegree[30];//保存入度
int graph[30][30];//是否有边
char output[30];//输出可确定序列
bool ok;//可以被确定
bool dilemma;//有环,矛盾
bool no;//不能被确定
char c1,c,c2;//输入

int topo(int n)
{
    int in[30];
    for(int i=0;i<n;i++)
        in[i]=indegree[i];//使用备用数组进行拓扑排序

    stack<int>s;//入度为0的点进栈
    for(int i=0;i<n;i++)
        if(!in[i])
        s.push(i);

    bool flag=0;//栈里面入度为0的元素大于一个时,不确定的拓扑排序
    int cnt=0;//入度为0的元素个数,也是输出序列里面的个数
    while(!s.empty())
    {
        if((s.size())>1)
            flag=1;//不确定
        int first=s.top();
        s.pop();
        output[cnt++]=first+'A';//放入输出序列里面
        for(int i=0;i<n;i++)
            if(graph[first][i])//与入度为0的元素相连的元素
        {
            in[i]--;
            if(in[i]==0)
                s.push(i);//入栈
        }
    }
    if(cnt!=n)//如果没有环的话,序列里面的元素个数肯定等于输入的元素个数,就算在某个元素未输入之前,它的入度也初始化为0
        return 2;//有环
    else if(flag==1)//不确定的拓扑排序
        return -1;
    return 1;
}

int main()
{
    int n,m;
    while(cin>>n>>m&&(n||m))
    {
        ok=0;dilemma=0;no=0;
        memset(indegree,0,sizeof(indegree));
        memset(graph,0,sizeof(graph));
        for(int i=1;i<=m;i++)
        {
            cin>>c1>>c>>c2;//当出现矛盾或者通过一些条件可被确定序列,剩下的输入条件就不需要再处理了
            if(!ok&&!dilemma)//没有环,没有确定的拓扑排序,这里的拓扑排序必须输入的字母都有。比如输入ABCD  那么只有AB不是确定的
            {
                int t1=c1-'A';
                int t2=c2-'A';
                if(graph[t2][t1])//双向边,有环,出现矛盾
                {
                    cout<<"Inconsistency found after "<<i<<" relations."<<endl;
                    dilemma=1;//出现矛盾
                    continue;
                }

                if(!graph[t1][t2])
                {
                    graph[t1][t2]=1;
                    indegree[t2]++;//入度++
                }
                int ans=topo(n);//确定返回1,有环返回2
                if(ans==2)
                {
                    cout<<"Inconsistency found after "<<i<<" relations."<<endl;
                    dilemma=1;
                    continue;
                }
                if(ans==1)
                {
                    cout<<"Sorted sequence determined after "<<i<<" relations: ";
                    for(int k=0;k<n;k++)
                        cout<<output[k];
                    cout<<"."<<endl;
                    ok=1;
                }
            }
        }
        if(!ok&&!dilemma)
            cout<<"Sorted sequence cannot be determined."<<endl;
    }
    return 0;
}

    原文作者:拓扑排序
    原文地址: https://blog.csdn.net/sr_19930829/article/details/37883895
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