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

 

 1 #include<iostream>
 2 #include<string>
 3 #include<cstdio>
 4 #include<cstring>
 5 
 6 using namespace std;
 7 
 8 int deg[30],d[30],g[30][30],res[30];
 9 int n,m;
10 
11 int Topo(string str){
12     int x=str[0]-'A';
13     int y=str[2]-'A';
14     if(g[x][y]==0){
15         g[x][y]=1;
16         d[y]++;
17     }
18     for(int i=0;i<n;i++)
19         deg[i]=d[i];
20     int flag=1,len=0,tmp,cnt;
21     for(int i=0;i<n;i++){
22         cnt=0;
23         for(int j=0;j<n;j++)
24             if(deg[j]==0){
25                 tmp=j;
26                 cnt++;
27             }
28         if(cnt==0)
29             return -1;
30         else if(cnt>1)      //这里不直接return 0;因为之后可能出现回路
31             flag=0;
32         deg[tmp]--;
33         res[len++]=tmp;
34         for(int j=0;j<n;j++)
35             if(g[tmp][j]==1)
36                 deg[j]--;
37     }
38     return flag;
39 }
40 
41 void Solve(){
42     string str;
43     int flag=1;
44     for(int i=0;i<m;i++){
45         cin>>str;
46         if(flag){
47             int tmp=Topo(str);
48             if(tmp==1){
49                 cout<<"Sorted sequence determined after "<<i+1<<" relations: ";
50                 for(int j=0;j<n;j++)
51                     cout<<char(res[j]+'A');
52                 cout<<"."<<endl;
53                 flag=0;
54             }else if(tmp==-1){
55                 cout<<"Inconsistency found after "<<i+1<<" relations."<<endl;
56                 flag=0;
57             }
58         }
59     }
60     if(flag)
61         cout<<"Sorted sequence cannot be determined."<<endl;
62 }
63 
64 int main(){
65 
66     //freopen("input.txt","r",stdin);
67 
68     string str;
69     while(scanf("%d%d",&n,&m)){
70         if(n==0 && m==0)
71             break;
72         if(n-1>m){
73             for(int i=0;i<m;i++)
74                 cin>>str;
75             cout<<"Sorted sequence cannot be determined."<<endl;
76             continue;
77         }
78         memset(g,0,sizeof(g));
79         memset(d,0,sizeof(d));
80         Solve();
81     }
82     return 0;
83 }