nyoj 349 Sorting It All Out（拓扑排序度的理解）

Sorting It All Out

时间限制：3000 ms  |  内存限制：65535 KB

难度：3

描述

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.

输入

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.

输出

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. 

样例输入

4 6A<BA<CB<CC<DB<DA<B3 2A<BB<A26 1A<Z0 0

样例输出

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

来源

POJ

上传者

陈玉

输入n和m，n表示26个字母前n个字母，m表示有多少个关系，然后输入m个关系，判断是否这n个字母存在一个排序关系
如果存在输出在几个关系之后就输出几个关系之后就可以确定，比如第一个测试数据，前四个关系输入之后，就输出结果
后两个关系输入不用管，，如果存在环那么就输出冲突，如果不能确定次序就输出不能确定。
拓扑排序的理解，度的判断 

#include<stdio.h>#include<string.h>#define M 30int re[M],path[M][M],du[M],d[M];char str[8];int topo(char s[],int n){int i,j,k;int A=s[0]-'A';int B=s[2]-'A';if(path[A][B]==0){path[A][B]=1;d[B]++;//因为每一次调用函数的时候要保持原来的图的数据，所以用d数组保存度不变 }for(i=0;i<n;i++)du[i]=d[i];//du数组来判断当前函数度的变化 int flag=1,cnt,pos=0;for(i=0;i<n;i++){cnt=0;for(j=0;j<n;j++){if(du[j]==0){cnt++;k=j;}}if(cnt==0) return -1;//cnt等于0表示存在环，即存在冲突 else if(cnt>1)flag=0;//存在多个度为0的点，即次序不能确定 du[k]--;re[pos++]=k;for(j=0;j<n;j++){if(path[k][j]==1)du[j]--;}}if(flag) return 1;return 0; }void result(int n,int m){int i,j;int ok=1;for(i=0;i<m;i++){scanf("%s",str);if(ok){int t=topo(str,n);if(t==1){printf("Sorted sequence determined after %d relations: ",i+1);for(j=0;j<n;j++)printf("%c",char(re[j]+'A'));printf(".\n");ok=0;}else if(t==-1){printf("Inconsistency found after %d relations.\n",i+1);ok=0;}}}if(ok)printf("Sorted sequence cannot be determined.\n");}int  main(){int n,m;char str[8];while(scanf("%d%d",&n,&m),n+m){if(m<n-1){//关系不够不能确定次序 for(int i=0;i<m;i++)scanf("%s",str);printf("Sorted sequence cannot be determined.\n");}else{memset(d,0,sizeof(d));memset(path,0,sizeof(path));result(n,m);}}return 0;}