poj1094Sorting It All Out

题目链接：

啊哈哈，选我

题目：

Sorting It All Out

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 26897 Accepted: 9281

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

这个题目对拓扑排序考虑的非常仔细。。

考虑了成环的情况。成环的情况也就是最后存在入度不为0的点。。则计数后最后的num不等于n。。

这个题目还有就是这个题目不是对所有的信息进行综合判断，而是根据前面的如果能够得到已经成环了，或者可以得到n的大小顺序了，则后面的就不用判断了。。。所以用ok1，ok2两个变量进行控制。。。

代码为：

#include<cstdio>#include<stack>#include<iostream>#include<cstring>#include<algorithm>using namespace std;const int maxn=26+10;int n,m;int in[maxn],Copy[maxn],map[maxn][maxn],temp[maxn];stack<int>S;int topo(){    int flag=0,num=0;    while(!S.empty())  S.pop();    memcpy(Copy,in,sizeof(in));    for(int i=0;i<n;i++)    {        if(Copy[i]==0)            S.push(i);    }    while(!S.empty())    {        if(S.size()>1)            flag=1;        int Gery=S.top();        S.pop();        temp[num++]=Gery;        for(int i=0;i<n;i++)        {            if(map[Gery][i])            {                if(--Copy[i]==0)                   S.push(i);            }        }    }    if(num!=n)        return 0;//成环，则已经可以确定关系了，可以标记。    if(flag)        return 1;//有多个入度为0的点，则还不确定，继续输入信息，增加条件，看是否能够得到顺序。    return 2;//顺序已经得到确定。可以标记。}int main(){    char str[maxn];    int ok1,ok2,u,v,i,is_n;    while(~scanf("%d%d",&n,&m),n,m)    {        is_n=0;        memset(in,0,sizeof(in));        memset(map,0,sizeof(map));        ok1=ok2=0;        for(i=1;i<=m;i++)        {            scanf("%s",str);            if(!ok1&&!ok2)            {                u=str[0]-'A';                v=str[2]-'A';                if(map[u][v]==0)                {                    map[u][v]=1;                    in[v]++;                }                int ans=topo();                if(ans==0)                {                    is_n=i;                    ok2=1;                }                else if(ans==2)                {                    is_n=i;                    ok1=1;                }            }        }        if(ok1)        {            printf("Sorted sequence determined after %d relations: ",is_n);            for(int i=0;i<n-1;i++)                printf("%c",temp[i]+'A');            printf("%c.\n",temp[n-1]+'A');        }        if(ok2)            printf("Inconsistency found after %d relations.\n",is_n);        if(ok1==0&&ok2==0)          printf("Sorted sequence cannot be determined.\n");    }    return 0;}