题目链接 http://poj.org/problem?id=1094
题目
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.
思路 分为3种情况，第一种是每个顶点的入度都不为0，这里就会有环路，无法形成拓扑排序。第二种情况是入度为了0的顶点不为1，这样就不能保证拓扑排序唯一性了。第三种情况就是保证入度为0的顶点只有一个，保证了拓扑排序的唯一性。
AC代码
#include<queue>
#include<vector>
#include<cstring>
#include<iostream>
#include<cstdio>
#include<string>
using namespace std;
int mapn[27][27]; //领接矩阵
int in[50],q[50]; // in表示入度，q存储排列
int toposort(int n)
{
int s1[50],flog=1,m,w,c;
c=0;
for(int i=0;i<n;i++)
s1[i]=in[i];
for(int i=0;i<n;i++){
m=0;
for(int j=0;j<n;j++){
if(s1[j]==0){
m++;
w=j;
}
}
if(m==0)
return 0;
if(m>1)
flog=-1;
q[c++]=w; // 将入度为0的点存储到q
s1[w]=-1; // 标记此点已用
for(int k=0;k<n;k++){
if(mapn[w][k]==1) // 如果w到k有通路
s1[k]--; // k点的入度减1
}
}
return flog;
}
int main()
{
int key,n,m;
while(cin>>n>>m){
if(n==0&&m==0)
break;
memset(mapn,0,sizeof(mapn));
memset(in,0,sizeof(in));
char s[5];
key=0; // 如果已经判断出来，则用key做标记
for(int i=1;i<=m;i++){
scanf("%s",s);
if(key)
continue;
int x=s[0]-'A';
int y=s[2]-'A';
mapn[x][y]=1;
in[y]++;
int w=toposort(n);
if(w==0){ // 第二种情况
printf("Inconsistency found after %d relations.\n",i);
key=1;
}
else if(w==1){ // 第三种情况
printf("Sorted sequence determined after %d relations: ",i);
for(int j=0;j<n;j++)
printf("%c",q[j]+'A');
printf(".\n");
key=1;
}
}
if(!key) // 第一种情况
printf("Sorted sequence cannot be determined.\n");
}
return 0;
}