POJ1094 Sorting It All Out 拓扑排序判大小关系

Sorting It All Out

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 36371 Accepted: 12800

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.

Source

East Central North America 2001

#include<stdio.h>#include<string.h>#include<vector>#include<iostream>using namespace std;int n,m,cnt;int degree[30];bool hp;bool edges[30][30];char topo[30];vector<char>vec;int toposort(){    vec.clear();    int indegree[30];//函数内部设一个记录入度的数组，尝试着进行拓扑;    int i,j,k,m;    memcpy(indegree, degree, sizeof(degree));    cnt = 0;    m = 0;//判断是否有多个前驱结点，注意初始化放在外循环外面；    bool insis=false;    for(i = 0; i < n; i++)    {        bool flag=true;        for(j = 0; j < n; j++)        {            if(indegree[j] == 0)            {                flag=false;                vec.push_back(j+'A');//进拓扑序列；                indegree[j]--;                for(k = j+1; k < n; k++)                {                    if(indegree[k] == 0)                        insis=true;//m=1说明有多个前驱结点，拓扑序列不唯一；                }                for(k = 0; k < n; k++)                {                    if(edges[j][k])                        indegree[k]--;//与j相连的结点入度减一；                }                break;            }        }        if (flag) return -1;    }    if (insis) return 0;    return 1;}int main(){int i,res;char u,v;while(~scanf("%d %d",&n,&m) && m && n){    memset(edges,0,sizeof(edges));    memset(degree,0,sizeof(degree));    hp=false;    getchar();        for(i = 1; i <= m; i++)        {            string rbs;            if (hp) {getline(cin, rbs); continue;}            scanf("%c<%c",&u,&v);            getchar();            if (edges[u-'A'][v-'A']==0){                degree[v-'A']++;                edges[u-'A'][v-'A'] = 1;            }            res = toposort();            if(res == -1)            {                //有环时直接输出                printf("Inconsistency found after %d relations.\n",i);                hp=true;            }            else if(res == 1)            {   //当只有一个前驱结点时才输出拓扑序列。                printf("Sorted sequence determined after %d relations: ",i);                for (int i=0; i<vec.size(); i++) printf("%c",vec[i]);                printf(".\n");                hp=true;            }        }if (!hp) printf("Sorted sequence cannot be determined.\n");}return 0;}

不能发现有多个0入度就直接return，因为可能会inconsistent！