Description
Bessie has been playing with strings again. She found that by changing the order of the alphabet she could make some strings come before all the others lexicographically (dictionary ordering). For instance Bessie found that for the strings "omm", "moo", "mom", and "ommnom" she could make "mom" appear first using the standard alphabet and that she could make "omm" appear first using the alphabet "abcdefghijklonmpqrstuvwxyz". However, Bessie couldn't figure out any way to make "moo" or "ommnom" appear first. Help Bessie by computing which strings in the input could be lexicographically first by rearranging the order of the alphabet. To compute if string X is lexicographically before string Y find the index of the first character in which they differ, j. If no such index exists then X is lexicographically before Y if X is shorter than Y. Otherwise X is lexicographically before Y if X[j] occurs earlier in the alphabet than Y[j].
Input
* Line 1: A single line containing N (1 <= N <= 30,000), the number of strings Bessie is playing with.
* Lines 2..1+N: Each line contains a non-empty string. The total number of characters in all strings will be no more than 300,000. All characters in input will be lowercase characters 'a' through 'z'. Input will contain no duplicate strings.
Output
* Line 1: A single line containing K, the number of strings that could be lexicographically first.
* Lines 2..1+K: The (1+i)th line should contain the ith string that could be lexicographically first. Strings should be output in the same order they were given in the input.
题解： 比较简单的一道题吧，考试的时候大家都几乎一眼切.
将所有串插进字典树
依次枚举每个串，将其定义为字典序最小的串.
那么，就要满足对于字典树中每一个分叉，必须保证当前串该字符大小要小于该分叉其余所有字符.
那么，这样就构建出了一些大小关系.
如果大小关系出现了环，说明无解.
我是用 tarjan 找环来判断的.
#include<bits/stdc++.h>
#define maxn 1000003
using namespace std;
char str[maxn],strtot[maxn];
int lentot,scc,flag,pp=0;
int st[maxn],ed[maxn],C[30][30],vis[maxn],vised[maxn],pre[maxn],low[maxn];
void setIO(string s)
{
string in=s+".in";
string out=s+".out";
freopen(in.c_str(),"r",stdin);
freopen(out.c_str(),"w",stdout);
}
vector<int>G[maxn],answer,tag[maxn];
stack<int>S;
struct Trie
{
int cnt;
int ch[maxn][30];
void ins(char p[],int o)
{
int len=strlen(p+1),cur=0;
for(int i=1;i<=len;++i)
{
int c=p[i]-'a';
if(!ch[cur][c])
{
ch[cur][c]=++cnt;
G[cur].push_back(c);
}
cur=ch[cur][c];
}
tag[cur].push_back(o);
}
}trie;
void tarjan(int u)
{
S.push(u);
vised[u]=1;
pre[u]=low[u]=++scc;
for(int i=0;i<27;++i)
{
if(u==i || !vis[i] || !C[u][i]) continue;
if(!vised[i]) tarjan(i), low[u]=min(low[u], low[i]);
else if(vised[i]==1) low[u]=min(low[u],pre[i]);
}
if(low[u]==pre[u])
{
int cc=0;
for(;;)
{
int x=S.top();S.pop();
++cc;
vised[x]=-1;
if(cc>1) flag=1;
if(x==u) break;
}
}
}
bool check()
{
flag=scc=0;
while(!S.empty())S.pop();
for(int i=0;i<27;++i) vised[i]=low[i]=pre[i]=0;
for(int i=0;i<27;++i)
{
if(!vis[i]) continue;
if(!vised[i]) tarjan(i);
}
return flag^1;
}
void dfs(int u,int depth)
{
if(tag[u].size())
{
if(check())
{
for(int j=0,sz=tag[u].size();j<sz;++j) answer.push_back(tag[u][j]);
}
return;
}
for(int i=0;i<27;++i)
{
// 有延申出去的单词
if(trie.ch[u][i]) // 当前为 u
{
++vis[i];
int y=u;
for(int j=0,sz=G[u].size();j<sz;++j)
{
if(G[u][j]!=i)
{
++C[i][G[u][j]];
}
}
dfs(trie.ch[u][i],depth+1);
for(int j=0,sz=G[u].size();j<sz;++j)
{
if(G[u][j]!=i) --C[i][G[y][j]];
}
--vis[i];
}
}
}
int main()
{
// setIO("ok");
int n;
scanf("%d",&n);
for(int i=1;i<=n;++i)
{
scanf("%s",str+1);
trie.ins(str,i);
int v=strlen(str+1);
st[i]=lentot+1;
for(int j=1;j<=v;++j) strtot[++lentot]=str[j];
ed[i]=lentot;
}
dfs(0,0);
printf("%d
",answer.size());
sort(answer.begin(),answer.end());
for(int i=0,sz=answer.size();i<sz;++i)
{
for(int j=st[answer[i]];j<=ed[answer[i]];++j) printf("%c",strtot[j]);
printf("
");
}
return 0;
}