POJ1094 拓扑排序

Sorting It All Out
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 31001 Accepted: 10766

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.

题意：
给出一个图里字母的先后顺序，若能进行拓扑排序，则在那一步输出排序结果，如果成环，则无法排序。如果没有用到所以字母，也是不能排序。
题解：
拓扑很好写，用个数组记录入度即可。问题是这个题的分类。它是只要在过程中能排序排序好了就可以输出了。成环也可以直接输出了。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
#include <queue>
#include <vector>
#define f(i,a,b) for(i = a;i<=b;i++)
#define fi(i,a,b) for(i = a;i>=b;i--)

using namespace std;

int InE[30];
int G[30][30];
char ss[30];
int n,m,Ti,S;

int Topo(){
    int cnt = 0;
    int tem[30] = {0},fla = 3;
    int i,j,S;
    f(i,1,n) tem[i] = InE[i];
    f(i,1,n){
        int sum = 0;
        f(j,1,n)
            if(!tem[j]){sum++;S = j;}
        if(sum == 0) return 1;
        if(sum >1)  fla = 2;
        ss[cnt++] = 'A' + S - 1;
        tem[S] = -1;
        f(j,1,n)
            if(G[S][j]) tem[j]--;
    }
    ss[n] = '\0';
    return fla;
}

int main()
{

    while(~scanf("%d%d",&n,&m)&&(n||m)){
        getchar();
        char a,b;
        int i,j,flag = 0,ok = 0;
        memset(G,0,sizeof(G));
        memset(InE,0,sizeof(InE));
        f(i,1,m){
            scanf("%c<%c",&a,&b);
            getchar();
            if(ok) continue;
            G[a-'A'+1][b-'A'+1] = 1;
            InE[b-'A'+1]++;
            flag = Topo();
            if(1 == flag){
                printf("Inconsistency found after %d relations.\n",i);
                ok = 1;
            }
            else if(3 == flag){
                printf("Sorted sequence determined after %d relations: %s.\n",i,ss);
                ok = 1;
            }
        }
        if(!ok)
            printf("Sorted sequence cannot be determined.\n");
    }
    return 0;
}