nyoj 349&poj 1094 Sorting It All Out（拓扑排序）

拓扑排序：若G包含有向边(U,V)，则在序列中U出现在V之前，即该序列使得图中所有有向边均从左指向右。如果图是有回路的，就不存在这样的序列。
　　首先选择一个无前驱的顶点(即入度为0的顶点，图中至少应该有一个这样的顶点，否则肯定存在回路)，然后从图中移去该顶点以及由其发出的所有有向边，如果图中还存在无前驱的顶点，则重复上述操作，直到操作无法进行。如果图不为空，说明图中存在回路，无法进行拓扑排序；否则移出的顶点的顺序就是对该图的一个拓扑排序。
具体思路：
　　每输入一组偏序关系进行一次拓扑排序。
　　如果存在回路，输出矛盾。
　　在不存在回路的基础上，判断每次入度为0的点是否唯一，只有保证每次只有一个点入度为0，才能保证最终的序列唯一。
注意：如果对于某一次输入已经能确定序列矛盾或者序列完全有序，则可以忽略后面的输入。

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 6A<BA<CB<CC<DB<DA<B3 2A<BB<A26 1A<Z0 0

Sample Output

Sorted sequence determined after 4 relations: ABCD.Inconsistency found after 2 relations.Sorted sequence cannot be determined.

#include<cstdio>#include<queue>#include<cstring>#include<vector>#include<stdlib.h>using namespace std;#define maxv 27#define maxe 101int n,m,size,head[maxv],indegree[maxv],ans,indegree1[maxv];vector<int>vv;struct edge{    int v,next;}edges[maxe];void init(){    size=0;    memset(head,-1,sizeof(head));memset(indegree,0,sizeof(indegree));}void insert(int u,int v){    edges[size].v=v;    edges[size].next=head[u];    head[u]=size++;    indegree[v]++;}int tsort(){    int i,flag=0;    queue<int>q;    vv.clear();    memcpy(indegree1,indegree,n*sizeof(int));//注意这里是对indegree1操作，不影响indegree，也就是用indegree1代替indegree操作，因为indegree要减1，会影响下一步    for(i=0;i<n;i++)        if(indegree1[i]==0)            q.push(i);    while(!q.empty())    {        if(q.size()>1) flag=1;        int w=q.front();        q.pop();        vv.push_back(w);                for(int i=head[w];i!=-1;i=edges[i].next)        {            indegree1[edges[i].v]--;            if(indegree1[edges[i].v]==0)                q.push(edges[i].v);        }    }    if(vv.size()!=n) return 1;    if(flag) return 0;    return 2;}int main(){    int i,j;    char s[4];    while(scanf("%d%d",&n,&m)&&n&&m)    {       init();       ans=0;       for(i=0;i<m&&!ans;i++)       {           scanf("%s",s);           int a=s[0]-'A',b=s[2]-'A';           insert(a,b);           ans=tsort();       }       for(j=i;j<m;j++) scanf("%s",s);       switch(ans)       {           case 0:printf("Sorted sequence cannot be determined.\n");break;           case 1:printf("Inconsistency found after %d relations.\n",i);break;           case 2:printf("Sorted sequence determined after %d relations: ",i);           for(i=0;i<n;i++)           {               printf("%c",vv[i]+'A');           }           printf(".\n");           break;       }    }    return 0;}