poj 1094（拓扑排序）

Language: Default Sorting It All Out Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 26943   Accepted: 9309 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

这是做的拓扑排序的第一题。

题意：输入n和m，n 表示给出的字符的个数，然后输入m行 字符之间的关系，每输入一行便要判断是不是有环，是不是有n个字符唯一的排列顺序。若全部输入完毕后没有唯一的排列顺序，也没有环便输出 Sorted sequence cannot be determined.

分析：每输入一行 字符间的关系便调用拓扑排序的函数判断，在拓扑排序中，先定义一个flag=1，如果有环，直接返回0；如果不唯一，则改变flag的值；有唯一情况时，保存入度为0的字符。循环完毕后，返回flag的值。

    在主函数中如果调用的函数返回1，那么表示有唯一排序，输出字符串；如果返回0，表示有环，停止循环。如果已输入m行，判断结果返回-1，那么输出 Sorted sequence cannot be determined.

#include <stdio.h>
#include <string.h>
int map[30][30];//记录图
int ans[30];//记录结果
int into[30];//记录入度
int into1[30];//同上
int n,m;

int topu()
{
	int i;
	int k;
	int flag=1;
	for(int j=1;j<=n;j++)//将入度保存到另一个数组中 ，确保into数组不变； 
	{
	  into1[j]=into[j];
	}
	for(i=1;i<=n;i++)
	{
		int count=0;
		for(int j=1;j<=n;j++)//找是否存在入度为0的点 
		{
			if(into1[j]==0)
			{
				 count++;
				 k=j;
			}
		}		
		if(count>1)flag= -1;//入度为0的点不唯一，所以顺序也不唯一 
		if(count==0)return 0;//不存在入度为0的点，有环； 
			
         ans[i]=k;
         into1[k]=-1;
           for(int j=1;j<=n;j++)//将从这个字符出发的到达其他的节点的入度减1； 
           {
              if(map[k][j]==1)
              	into1[j]--;
           }
		
	}
	return flag;
}
int main()
{
	while(scanf("%d%d",&n,&m),m,n)
	{
		memset(map,0,sizeof(map));//清空数组 
		memset(ans,0,sizeof(ans));
		memset(into,0,sizeof(into));
		char a[5];
		int i;
		int flag=0;
	    for(i=1;i<=m;i++)
		{
			scanf("%s",&a);
			if(flag)continue; //已经输出过了，所以continue 
			int x=a[0]-'A'+1;
			int y=a[2]-'A'+1;
			if(!map[x][y])//消除重边 
			{
				map[x][y]=1;
				into[y]++;
			}
			int ans1=topu();//调用拓扑排序 
				
			if(ans1==0)
			{
				printf("Inconsistency found after %d relations.\n",i);
				flag=1;
			}
			if(ans1==1)
			{
				printf("Sorted sequence determined after %d relations: ",i);
				for(int j=1;j<=n;j++)
					printf("%c",ans[j]+'A'-1);
				printf(".\n");
				flag=1;
			}
		}
		if(!flag)//表示没有输出过 
			printf("Sorted sequence cannot be determined.\n");
	}
	return 0;
}