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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.
InputInput 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.
OutputFor 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 Input4 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 OutputSorted sequence determined after 4 relations: ABCD.
Inconsistency found after 2 relations. Sorted sequence cannot be determined.题意是给出一些比较关系，求给出这些关系的过程中1 ）能否有一个确定的排序，有的话给出关系的个数并求出这个排序的序列；2）是否成环；3）不能确定
只有一个确定排序的拓扑序列的性质应该是，只有一个入度为0的节点，去掉此节点后，剩下的还是只有一个入度为0的点。按照这个性质模拟就行了。#include#include #include #include #include #include #include #include #include #define INF 0x3f3f3f3f#define MAXN 105#define Mod 10001using namespace std;int n,m,c;int map[30][30],temp[30],indegree[30],q[30];int topsort(int n){ int m,pos,flag=1; c=0; for(int i=1; i<=n; ++i) temp[i]=indegree[i]; for(int i=1; i<=n; ++i) { m=0; for(int j=1; j<=n; ++j) if(temp[j]==0) //选择一个入度为0的点，即可能为起点 { m++; pos=j; } if(m==0) return 0; //有环 if(m>1) flag=-1; //还不确定 q[c++]=pos; //该起点入栈 temp[pos]=-1; for(int j=1;j<=n;++j) if(map[pos][j]) //把起点去掉后，相连的点要有处理 temp[j]--; } return flag;}int main(){ char op[10]; while(~scanf("%d%d",&n,&m)) { getchar(); if(n==0&&m==0) break; memset(indegree,0,sizeof(indegree)); memset(map,0,sizeof(map)); int s=-1,flag=0; for(int i=1; i<=m; ++i) { gets(op); if(s==0||s==1) continue; int x=op[0]-'A'+1; int y=op[2]-'A'+1; map[x][y]=1; indegree[y]++; s=topsort(n); if(s==0) { printf("Inconsistency found after %d relations.\n",i); flag=1; } if(s==1) { printf("Sorted sequence determined after %d relations: ",i); for(int j=0;j
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