Sorting It All Out （拓扑排序）

Problem 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: <br> <br>Sorted sequence determined after xxx relations: yyy...y. <br>Sorted sequence cannot be determined. <br>Inconsistency found after xxx relations. <br> <br>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. <br>

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.

题目大概：

看是否成立大小关系，分别对应三种输出关系。

思路：

这道题，是关于比大小的，一般用拓扑排序是解决的是有向无环图，当然也可以检验出是否有环，按照题目上说，有三种输出，一种是组成了大小关系，即形成了环。还有自相矛盾，不能形成大小关系。最后一种是给出条件不能判断关系，由于输出的不同，所以也不能按常理来做这道拓扑排序题，这道题的大小数据应该输入一个，判断一个，一旦发现条件成立就输出，因为输出条件要包含，在哪一个输入下，满足的条件，这就相当于把拓扑排序最外面那层循环和输入条件的循环合并了，再就是控制条件比较多，需要加一些变量来控制输出。

代码：

借鉴自大佬。

#include <iostream>#include<cstdio>#include<cstring>#include<vector>#include<queue>using namespace std;int n,m,r[105],c[105],map[105],t,r1,num;char a, b, d;vector<int>w[105];queue<int>q;void init(){    memset(r, 0, sizeof(r));    for(int i=0; i<=n; ++i){        w[i].clear();    }} bool find(int u,int v){    for(int i=0; i<w[u].size(); ++i)        if(w[u][i]==v)return true;    return false;}int go(){    while(!q.empty())q.pop();    for(int i=0; i<n; ++i)if(r[i]==0){            q.push(i);    }    r1=0;    bool tt=false;    while(!q.empty()){        if(q.size()>1) tt=true;        int t=q.front();        q.pop();        map[r1++]=t;        for(int i=0; i<w[t].size(); ++i){            if(--r[w[t][i]]==0)                q.push(w[t][i]);        }    }    if(r1<n) return 1;    if(tt)  return 2;    return 3;}int main(){    int x,y,i,f,ok,stop;    while(cin>>n>>m){        if(!n||!m)break;        init();        f=2;        ok=false;        for(i=1; i<=m; ++i){            cin>>a>>b>>d;            if(ok) continue;            x=a-'A', y=d-'A';            if(b=='<'&&!find(y,x)){                   w[y].push_back(x);                    r[x]++;            }            else if(b=='>'&&!find(x,y)){                    w[x].push_back(y);                    r[y]++;            }            memcpy(c, r, sizeof(r));            f=go();            memcpy(r, c, sizeof(c));            if(f!=2){                stop=i;                ok=true;            }        }        if(f==3){            printf("Sorted sequence determined after %d relations: ", stop);            for(int i=r1-1; i>=0; --i)                printf("%c",map[i]+'A');            printf(".\n");        }        else if(f==1){            printf("Inconsistency found after %d relations.\n",stop);        }        else{            printf("Sorted sequence cannot be determined.\n");        }    }    return 0;}