poj 1094 Sorting It All Out

Sorting It All Out

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 27962 Accepted: 9666

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.

这道题目是拓扑排序，就是操作烦了点。这道题还有个坑爹的地方，就是一旦找到拓扑序列，直接输出答案，即使后面出现了矛盾。

首先，把n个点看成孤立的点，然后不停地加边（可以有平行边），每加一条边就做一次拓扑排序，刚开始并未出现矛盾，拓扑序列个数为n，但是每次拓扑单次向栈里加元素会超过一个，这表示关系未确定。

之后可能出现三种情况：矛盾（拓扑序列个数小于n），做个标记，以后只需把读入读完，不用做拓扑排序了。

找到序列（拓扑序列个数为n，并且单次入栈仅有1个元素），同样做个标志，以后只需把读入读完，不用做拓扑排序了。 没有找到序列并且不矛盾，（拓扑序列个数为n，并且某次入栈超过1个元素），继续读入数据并加边，做拓扑排序。

这样并且需要记录前几句话可以判断，只需在标志位改变同时，记录第几句话就可以了，然后每次用一个数组在拓扑排序时，记录拓扑序列，方便打印。

代码：

#include<cstdio>#include<iostream>#include<cstring>using namespace std;char s[4],ret[26];int adj[26][26],indeg[26],cp[26];int topo_sort(int n){    int top=-1,cnt=0,ans=0,tot=0;    memmove(cp,indeg,sizeof(int)*n);    for(int i=0;i<n;i++)        if(!cp[i]) cp[i]=top,top=i,cnt++;    int maxx=cnt;    while(top!=-1){        int j=top;        ret[tot++]='A'+j;        top=cp[top],cnt=0,ans++;        for(int i=0;i<n;i++)            if(adj[j][i])                if(!(cp[i]-=adj[j][i])) cp[i]=top,top=i,cnt++;        maxx=max(cnt,maxx);    }    ret[tot]='\0';    if(ans!=n) return 0;    return maxx;}int main(){    int n,m,res;    while(scanf("%d%d",&n,&m),n){        memset(adj,0,sizeof adj);        memset(indeg,0,sizeof indeg);        bool flag=true,first=true;        for(int i=1;i<=m;i++){            scanf("%s",s);            if(!flag||!first) continue;            int p=s[0]-'A',q=s[2]-'A';            adj[p][q]++;            indeg[q]++;            int maxx=topo_sort(n);            if(!maxx) res=i,flag=false;            else if(maxx==1&&first) res=i,first=false;        }        if(!flag) printf("Inconsistency found after %d relations.\n",res);        else if(first) puts("Sorted sequence cannot be determined.");        else printf("Sorted sequence determined after %d relations: %s.\n",res,ret);    }return 0;}