• POJ1094 Sorting It All Out 拓扑排序判断状态


    Sorting It All Out
    Time Limit: 1000MS   Memory Limit: 10000K
    Total Submissions: 21865   Accepted: 7529

    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.
    /*
    这道题WA了好久,其中有几个需要注意的地方
    1、当出现正好存在一种情况能够排序完所有节点时,不管以后的边会出现什么情况,都输出
        能够排序成功
    2、当中间的拓扑排序过程中出现多个几点的入度为0时,只记录当时的状态
        (亦即该测试数据要么出现环,要么就是有多组解),不能立即返回,
        要继续读边,直到能够排序完成(此时输出有多解的情况)或者出现环。
    */
    #include<cstdio>
    #include<iostream>
    #include<cstring>
    
    using namespace std;
    
    bool G[30][30];
    int d[30];
    char s[30];
    
    int toposort(int n)
    {
    	int num,k,i,j,t;
    	int td[30];
    	bool flag=true;
    	for(i=1;i<=n;++i)
    		td[i]=d[i];
    	memset(s,0,sizeof(0));
    	for(j=0;j<n;++j)
    	{
    		num=0;
    		for(i=1;i<=n;++i)
    		{
    			if(td[i]==0)
    			{
    				k=i;
    				++num;
    			}
    		}
    		if(num==0)    //有环
    			return -1;
    		if(num>1)  //有多种情况,还需继续读边判断
            {
                flag=false;
            }
    		s[j]='A'+k-1;
    		td[k]=-1;
    		for(t=1;t<=n;++t)
    		{
    			if(G[k][t])
    				--td[t];
    		}
    	}
    	s[n]='\0';
    	if(flag==false)  //情况不唯一
            return 0;
    	return 1;      //全部排好序了返回1.
    }
    
    int main()
    {
    	int n,m,i,ans,k;
    	bool flag;
    	char temp[5];
    	while(scanf("%d%d",&n,&m),m||n)
    	{
    		memset(G,false,sizeof(G));
    		memset(d,0,sizeof(d));
    		flag=true;
    		for(i=1;i<=m;++i)
    		{
    			scanf("%s",temp);
    			if(!flag)         //已经排好序或者有环
    				continue;
    			int u=temp[0]-'A'+1;
    			int v=temp[2]-'A'+1;
    			if(!G[u][v])
    			{
    				++d[v];
    				G[u][v]=true;
    			}
    			ans=toposort(n);
    			if(ans==1||ans==-1)
    			{
    				k=i;
    				flag=false;
    			}
    		}
    		if(ans==1)
    		{
    			printf("Sorted sequence determined after %d relations: %s.\n",k,s);
    		}
    		else if(ans==-1)
    		{
    			printf("Inconsistency found after %d relations.\n", k);
    		}
    		if(flag&&i>m)
    			printf("Sorted sequence cannot be determined.\n");
    	}
    	return 0;
    }
    

      




    功不成,身已退
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  • 原文地址:https://www.cnblogs.com/dongsheng/p/2749896.html
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