poj 1094 Sorting It All Out (拓扑排序)

Sorting It All Out

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 26088		Accepted: 9035

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 6
A<B
A<C
B<C
C<D
B<D
A<B
3 2
A<B
B<A
26 1
A<Z
0 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

题意：

    给出n个大写字母一系列的大小关系，问给出第几个时能确定其全部关系(是否确定全部大小)，关系有三种：有环矛盾、不确定、确定。

    这题有个小陷阱，就是要每输入一个关系就要topo判断一次，确定后之后的关系不影响，简单处理掉就行。

拓扑排序：

    这题其实不难，全部处理完如果不是确定关系或矛盾关系即是不确定关系。处理时难点好像也没有，就是注意一下in和V的复制。

 

//180K    16MS    C++    1629B    2014-04-11 17:30:45
#include<iostream>
#include<queue>
#include<vector>
using namespace std;
vector<int>V[30],tV[30];
int in[30];
int n,m;
int certain,conflict;
void topo(int times)
{
    queue<int>Q;
    char ans[30];
    memset(ans,0,sizeof(ans));
    int sum=0,uncertain=0;
    int tin[30];
    for(int i=0;i<n;i++){ //复制 
        tin[i]=in[i];
        tV[i].resize(V[i].size());
        memcpy(&tV[i][0],&V[i][0],V[i].size()*sizeof(int));
    }
    for(int i=0;i<n;i++){
        if(tin[i]==0){
            Q.push(i);tin[i]--;
        }
    }
    
    while(!Q.empty()){
        if(Q.size()>1) uncertain=1;
        int t=Q.front();
        Q.pop();
        ans[sum++]=t+'A';
        tin[t]--;
        for(int i=0;i<tV[t].size();i++){
            if(--tin[tV[t][i]]==0){
                Q.push(tV[t][i]);
            }
        }
    }
    //printf("**%d\n",sum); 
    if(sum<n){
        conflict=1;printf("Inconsistency found after %d relations.\n",times);
    }
    else if(uncertain) return;
    else{
        certain=1;
        printf("Sorted sequence determined after %d relations: %s.\n",times,ans);
    }
}
int main(void)
{
    char s[5];
    while(scanf("%d%d",&n,&m),n+m)
    {
        for(int i=0;i<30;i++){
            in[i]=0;
            V[i].clear();
        }
        certain=conflict=0;
        for(int i=0;i<m;i++){
            scanf("%s",s);
            V[s[0]-'A'].push_back(s[2]-'A');
            in[s[2]-'A']++;
            if(!certain && !conflict) topo(i+1);
        }
        if(!certain && !conflict) puts("Sorted sequence cannot be determined.");
    }
}