Codeforces 510C Fox And Names 拓扑排序
Fox Ciel is going to publish a paper on FOCS (Foxes Operated Computer Systems, pronounce: "Fox"). She heard a rumor: the authors list on the paper is always sorted in the lexicographical order.
After checking some examples, she found out that sometimes it wasn‘t true. On some papers authors‘ names weren‘t sorted inlexicographical order in normal sense. But it was always true that after some modification of the order of letters in alphabet, the order of authors becomes lexicographical!
She wants to know, if there exists an order of letters in Latin alphabet such that the names on the paper she is submitting are following in the lexicographical order. If so, you should find out any such order.
Lexicographical order is defined in following way. When we compare s and t, first we find the leftmost position with differing characters: si?≠?ti. If there is no such position (i. e. s is a prefix of t or vice versa) the shortest string is less. Otherwise, we compare characters si and tiaccording to their order in alphabet.
The first line contains an integer n (1?≤?n?≤?100): number of names.
Each of the following n lines contain one string namei (1?≤?|namei|?≤?100), the i-th name. Each name contains only lowercase Latin letters. All names are different.
If there exists such order of letters that the given names are sorted lexicographically, output any such order as a permutation of characters ‘a‘–‘z‘ (i. e. first output the first letter of the modified alphabet, then the second, and so on).
Otherwise output a single word "Impossible" (without quotes).
3 rivest shamir adleman
bcdefghijklmnopqrsatuvwxyz
10 tourist petr wjmzbmr yeputons vepifanov scottwu oooooooooooooooo subscriber rowdark tankengineer
Impossible
10 petr egor endagorion feferivan ilovetanyaromanova kostka dmitriyh maratsnowbear bredorjaguarturnik cgyforever
aghjlnopefikdmbcqrstuvwxyz
7 car care careful carefully becarefuldontforgetsomething otherwiseyouwillbehacked goodluck
acbdefhijklmnogpqrstuvwxyz
解题方法：先把字典字母之间的关系找到。通过图连接起来。然后用拓扑排序
如果拓扑排序还没有学过。可以看http://www.cnblogs.com/newpanderking/archive/2012/10/18/2729552.html 这里面讲的很好
#include<iostream>
#include<string>
#include<algorithm>
#include<cstdlib>
#include<cstdio>
#include<set>
#include<map>
#include<vector>
#include<cstring>
#include<stack>
#include<cmath>
#include<queue>
#define INF 0x0f0f0f0f
using namespace std;
char s[105][105],ans[30];
int cou[30];
bool link[30][30];
queue<int> q;
bool solve() //拓扑排序
{
int i,j,k,l,font,cnt=0;
for(i=0;i<26;i++)
{
if(!cou[i])
{
q.push(i);
ans[cnt++]='a'+i;
}
}
while(!q.empty())
{
font=q.front();
q.pop();
for(i=0;i<26;i++)
{
if(link[font][i])
{
cou[i]--;
if(cou[i]==0)
{
q.push(i);
ans[cnt++]='a'+i;
}
}
}
}
ans[26]='\0';
if(cnt<26) return false;
else return true;
}
int main()
{
int i,j,k,l,m,n,l1,l2,jud;
k=1;
memset(link,0,sizeof(link));
scanf("%d%*c",&n);
for(i=0;i<n;i++)
{
scanf("%s",&s[i]);
}
for(i=0;i<n-1;i++)
{
l1=strlen(s[i]);
l2=strlen(s[i+1]);
jud=0;
for(j=0;j<l1&&j<l2&&!jud;j++)
{
if(s[i][j]!=s[i+1][j])
{
jud=1;
if(!link[s[i][j]-'a'][s[i+1][j]-'a'])
{
link[s[i][j]-'a'][s[i+1][j]-'a']=true;
cou[s[i+1][j]-'a']++;
}
}
}
if(!jud&&l1>l2) k=0; //某两个字符串前缀完全相同，但是第一个字符串长度较大，则两字符串必定不满足字典序
}
if(!k)
{
printf("Impossible\n");
}
else
{
k=solve();
if(!k) printf("Impossible\n");
else printf("%s\n",ans);
}
return 0;
}
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