NYOJ 349 Sorting It All Out （拓扑排序 ）

题目链接

描述

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.

• 输入
  
  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.
• 输出
  
  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.
• 样例输入
  
  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
• 样例输出
  
  Sorted sequence determined after 4 relations: ABCD.
  
  Inconsistency found after 2 relations.
  
  Sorted sequence cannot be determined.

分析：

首先补充一下拓扑排序的思想：

（1）从有向图中选择一个没有前驱（入度为0）的顶点并输出它。

（2）从图中删除该节点，并且删去从该节点出发的全部有向边。

（3）重复上述操作，知道图中不在存在没有前驱的顶点为止。

这样操作的结果有两种：一种是图中全部定点被输出，这说明图中不存在有向回路；另一种是图中顶点未被全部输出，剩余的顶点均有前驱节点，这说明图中存在有向回路。

这就是一个典型的拓扑排序的应用，如果能够排好序的话，说明就是可以的，如果形成环的话，说明能够构成回路也就是存在冲突，否则就是不能够排好序。

代码：

#include<stdio.h>
#include<iostream>
#include<vector>
#include<queue>
#include<string.h>
using namespace std;
int du[30];
char ch[30];
int n,m,k;
vector<int> v[30];
int init()
{
    memset(du,0,sizeof(du));
    memset(v,0,sizeof(v));
}
int topSort()
{
    int op=1;
    k=0;
    queue<int >q;
    int du1[30];
    for(int i=0; i<n; i++)///要把入度的函数复制一下，不然会影响下次的排序
    {
        du1[i]=du[i];
        if(du1[i]==0)
            q.push(i);
    }
    while(!q.empty())
    {
        //cout<<"  ----"<<q.size()<<endl;
        if(q.size()>1) op=0;///相当于有多个入度为0的点，也就是还没有排好序
        int a=q.front();
        q.pop();
        char c=a+'A';
        ch[k++]=c;
        // cout<<"k="<<k<<endl;
        for(int i=0; i<v[a].size(); i++)
        {
            int b=v[a][i];
            du1[b]--;
            if(du1[b]==0)
                q.push(b);
        }
    }
    if(k<n)///形成环，如果没有形成环且能排好序的话，肯定每一个点都要入队一次
        return -1;
    return op;///op==1，已经排好序；op=1，还没有排好
}
int main()
{
    char ch2,ch1;
    int a,b;
    while(~scanf("%d%d",&n,&m),n,m)
    {
        init();
        int flag=0;
        for(int mm=1; mm<=m; mm++)
        {
            scanf(" %c<%c",&ch1,&ch2);
            if(flag!=0) continue;///已经确定当前的序列是有序或者已经发生冲突
            a=ch1-'A';
            b=ch2-'A';
            // cout<<a<<"   "<<b<<endl;
            v[a].push_back(b);///单向
            du[b]++;
            flag=topSort();///每次加入一个都要进行一次判断，看能否满足某个条件
            if(flag==1)///已经排好序
            {
                printf("Sorted sequence determined after %d relations: ",mm);
                for(int k1=0; k1<k; k1++)
                {
                    printf("%c",ch[k1]);
                }
                printf(".\n");
            }
            if(flag==-1)///发生冲突
            {
                printf("Inconsistency found after %d relations.\n",mm);
            }
        }
        if(flag==0)///到最后还没有排好序
        {
            printf("Sorted sequence cannot be determined.\n");
        }
    }
    return 0;
}