POJ 1094 Sorting It All Out【拓扑排序】

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.

题目大意：

       给定前 n 个字母，以及一些他们之间的大小关系，求这些字母满足大小关系的一个序列；

       在大小关系给定的过程中，如果已经可以满足唯一有序，则可以按顺序输出序列；

       如果现有关系是相互矛盾的，则输出发生矛盾的关系的序号；

       如果最终关系满足有序，但不满足唯一有序，则输出 序列不能够确定。

解题思路：

      一开始居然在思考 一笔画问题，后来发现，关系出现矛盾，即在拓扑排序中发现环，才立刻转上了拓扑排序的车。

     这个的输入数据可以不完全采纳，感觉有点麻烦，加一个标记变量即可。

     则，每一组新的关系加入，就进行一次拓扑排序：【拓扑排序：https://blog.youkuaiyun.com/sodacoco/article/details/86586881】

            1》如果出现环，则可以结束，关系相互矛盾；

            2》如果已经可以唯一有序，则可以结束，输出这个序列；

            3》所有关系都已经输入，还未产生以上两种结果，则关系无法确定；

代码实现：

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int Map[30][30],in[30],que[30];

int topu(int n){
    int ru0,dian,flag=1;
    int tempIn[30],queNum=0;
    for(int i=1;i<=n;i++)
        tempIn[i]=in[i];

    for(int i=1;i<=n;i++){
        ru0=0;
        for(int j=1;j<=n;j++){
            if(tempIn[j]==0){
                ru0++; dian=j;  //下一个入栈的点
            }
        }
        if(ru0==0) return 0;    //对于现阶段的子图来说存在环
        if(ru0>1) flag=-1;      //对于现阶段的子图来说，有多个入度为0的点

        que[++queNum]=dian;     //入数组
        tempIn[dian]=-1;
        for(int j=1;j<=n;j++)
            if(Map[dian][j]==1)
                tempIn[j]--;
    }
    return flag;
}
int main(){
    int n,m,biao=0;
    char ch1,ch2;
    while(scanf("%d%d",&n,&m)&&(n+m)!=0){
        getchar();
        memset(Map,0,sizeof(Map));
        memset(in,0,sizeof(in));
        biao=0;
        for(int i=1;i<=m;i++){
            ch1=getchar();  getchar();
            ch2=getchar();  getchar();
            if(biao==1)continue;

            int x=ch1-'A'+1;
            int y=ch2-'A'+1;
            Map[x][y]=1;  in[y]++;

            int ans=topu(n);
            if(ans==0){ //有环
                printf("Inconsistency found after %d relations.\n",i);
                biao=1;
            }
            if(ans==1){ //有序
                printf("Sorted sequence determined after %d relations: ",i);
                for(int j=1;j<=n;j++)   printf("%c",que[j]+'A'-1);
                printf(".\n");
                biao=1;
            }
        }
       if(!biao) //不确定
            printf("Sorted sequence cannot be determined.\n");
    }
    return 0;
}