POJ 1094 Sorting It All Out
2013-04-12 12:07
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Sorting It All Out
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 |
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;
}
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