POJ - 1094 Sorting It All Out

Sorting It All Out Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 35720   Accepted: 12569 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

nyoj上也有这道题，但是数据比较水。

拓扑排序。每输入一行都要进行拓扑排序。需要判断是否存在环，是否已经排序成功。

判定环：利用Kahn算法进行删点后，如果还存在入度大于0的点，则存在环。

判定排序成功：如果每次只能删一个点，且没有环，则排序成功。

如果有环，或者排序成功，就不需要继续拓扑排序了。

代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<queue>
#include<vector>
using namespace std;
int n,m,in[30],buf[30];
vector<int> vec[30];
int topo()
{
    int degree[30],ok=1,num=0,cnt=0;
    memcpy(degree,in,sizeof in);
    queue<int> q;
    for(int i=0;i<n;i++)
    {
        if(degree[i]==0)
        {
            cnt++;
            q.push(i);
        }
        if(cnt>1)
            ok=0;
    }
    while(!q.empty())
    {
        int top=q.front(); q.pop();
        buf[num++]=top;
        cnt=0;
        for(int i=0;i<vec[top].size();i++)
        {
            int now=vec[top][i];
            degree[now]--;
            if(!degree[now])
            {
                cnt++;
                q.push(now);
            }
        }
        if(cnt>1)//未完成
            ok=0;
    }
    for(int i=0;i<n;i++)
        if(degree[i])
            return -1;//有环
    return ok;
}
int main()
{
    char op[5];
    while(scanf("%d%d",&n,&m) && n+m)
    {
        int flag=0;
        for(int i=0;i<n;i++)
            vec[i].clear();
        memset(in,0,sizeof in);
        for(int i=1;i<=m;i++)
        {
            scanf("%s",op);
            in[op[2]-'A']++;
            vec[op[0]-'A'].push_back(op[2]-'A');
            if(flag==1 || flag==-1)
                continue;
            flag=topo();
            if(flag==1)
            {
                printf("Sorted sequence determined after %d relations: ",i);
                for(int i=0;i<n;i++)
                    printf("%c",buf[i]+'A');
                printf(".\n");
            }
            if(flag==-1)
                printf("Inconsistency found after %d relations.\n",i);
        }
        if(flag==0)
            printf("Sorted sequence cannot be determined.\n");
    }
    return 0;
}
/*
 6 5
 A<D
 B<C
 A<B
 D<E
 C<D

6 6
 A<D
 B<C
 E<F
 B<A
 D<E
 A<B

 6 6
 D<E
 B<F
 A<D
 F<C
 A<F
 E<B

 3 3
 A<B
 B<C
 C<A

 4 6
 A<B
 A<C
 B<C
 C<D
 B<D
 A<B

 3 2
 A<B
 B<A

 2 2
 A<B
 B<A

 26 1
 A<Z

 0 0
*/