POJ 1094 Sorting It All Out

Sorting It All Out

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 23082		Accepted: 7978

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

题目大意：
　　给出一些字母和它们之间的偏序关系，让你来判断通过这些关系是否能构成唯一的升序序列。如果能，输出这个序列，并输出是通过前多少关系得出的（即使之后的关系引出矛盾也不必理会）。如果存在矛盾，则输出前多少项出现的矛盾的。如果输入完仍无法得出唯一关系，输出相关信息。
题目算法：
　　拓扑排序。
　　拓扑排序：若G包含有向边(U,V)，则在序列中U出现在V之前，即该序列使得图中所有有向边均从左指向右。如果图是有回路的，就不存在这样的序列。
　　首先选择一个无前驱的顶点(即入度为0的顶点，图中至少应该有一个这样的顶点，否则肯定存在回路)，然后从图中移去该顶点以及由其发出的所有有向边，如果图中还存在无前驱的顶点，则重复上述操作，直到操作无法进行。如果图不为空，说明图中存在回路，无法进行拓扑排序；否则移出的顶点的顺序就是对该图的一个拓扑排序。
具体思路：
　　每输入一组偏序关系进行一次拓扑排序。
　　如果存在回路，输出矛盾。
　　在不存在回路的基础上，判断每次入度为0的点是否唯一，只有保证每次只有一个点入度为0，才能保证最终的序列唯一。
注意：如果对于某一次输入已经能确定序列矛盾或者序列完全有序，则可以忽略后面的输入。

当入度为0的定点大于1时，直接return了，但之后可能出现回路即矛盾的情况。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<stack>

using namespace std;

int n,m,indeg[30],vis[30];
vector<int> vt[30];
char res[30];

int TopoSort(){
    stack<int> st;
    int flag=1;
    for(int i=0;i<n;i++){
        vis[i]=indeg[i];
        if(indeg[i]==0)
            st.push(i);
    }
    if(st.size()>1)
        flag=0;
    int cnt=0;
    while(!st.empty()){
        int u=st.top();
        st.pop();
        res[cnt++]=(char)u+'A';
        for(int i=0;i<vt[u].size();i++)
            if(--vis[vt[u][i]]==0)
                st.push(vt[u][i]);
        if(st.size()>1)
            flag=0;
    }
    res[cnt]='\0';
    for(int i=0;i<n;i++)
        if(vis[i])
            return -1;
    return flag && (cnt==n);
}

int main(){

    //freopen("input.txt","r",stdin);

    char str[5];
    int tag,flag,ans;
    while(~scanf("%d%d",&n,&m)){
        if(n==0 && m==0)
            break;
        memset(indeg,0,sizeof(indeg));
        for(int i=0;i<n;i++)
            vt[i].clear();
        ans=tag=0;
        for(int i=1;i<=m;i++){
            scanf("%s",str);
            vt[str[0]-'A'].push_back(str[2]-'A');
            indeg[str[2]-'A']++;
            if(tag==0){
                flag=TopoSort();
                if(flag==-1){
                    ans=i;
                    tag=-1;
                }
                if(flag==1){
                    ans=i;
                    tag=1;
                }
            }
        }
        if(tag==1)
            printf("Sorted sequence determined after %d relations: %s.\n",ans,res);
        else if(tag==-1)
             printf("Inconsistency found after %d relations.\n",ans);
        else if(tag==0)
             printf("Sorted sequence cannot be determined.\n");
    }
    return 0;
}

一些练习：http://www.cnblogs.com/372465774y/category/424141.html

#include<iostream>
#include<string>
#include<cstdio>
#include<cstring>

using namespace std;

int deg[30],d[30],g[30][30],res[30];
int n,m;

int Topo(string str){
    int x=str[0]-'A';
    int y=str[2]-'A';
    if(g[x][y]==0){
        g[x][y]=1;
        d[y]++;
    }
    for(int i=0;i<n;i++)
        deg[i]=d[i];
    int flag=1,len=0,tmp,cnt;
    for(int i=0;i<n;i++){
        cnt=0;
        for(int j=0;j<n;j++)
            if(deg[j]==0){
                tmp=j;
                cnt++;
            }
        if(cnt==0)
            return -1;
        else if(cnt>1)      //这里不直接return 0;因为之后可能出现回路
            flag=0;
        deg[tmp]--;
        res[len++]=tmp;
        for(int j=0;j<n;j++)
            if(g[tmp][j]==1)
                deg[j]--;
    }
    return flag;
}

void Solve(){
    string str;
    int flag=1;
    for(int i=0;i<m;i++){
        cin>>str;
        if(flag){
            int tmp=Topo(str);
            if(tmp==1){
                cout<<"Sorted sequence determined after "<<i+1<<" relations: ";
                for(int j=0;j<n;j++)
                    cout<<char(res[j]+'A');
                cout<<"."<<endl;
                flag=0;
            }else if(tmp==-1){
                cout<<"Inconsistency found after "<<i+1<<" relations."<<endl;
                flag=0;
            }
        }
    }
    if(flag)
        cout<<"Sorted sequence cannot be determined."<<endl;
}

int main(){

    //freopen("input.txt","r",stdin);

    string str;
    while(scanf("%d%d",&n,&m)){
        if(n==0 && m==0)
            break;
        if(n-1>m){
            for(int i=0;i<m;i++)
                cin>>str;
            cout<<"Sorted sequence cannot be determined."<<endl;
            continue;
        }
        memset(g,0,sizeof(g));
        memset(d,0,sizeof(d));
        Solve();
    }
    return 0;
}