【Codeforces Round #290 (Div. 2)-C. Fox And Names】 思維題+拓撲排序
阿新 • • 發佈:2018-11-14
Codeforces Round #290 (Div. 2)-C. Fox And Names
題意
做法
坑點
程式碼
#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;
const int maxn = 105;
char str[maxn][maxn];
vector<int> v[maxn];
int ind[maxn], que[maxn],m;//ind陣列要清空
bool topsort()//還要特判自環
{
int qt = 0;
for(int i = 0;i<26; ++i)
{
if(ind[i] == 0) que[qt++] = i;
}
for(int i = 0; i < qt; ++i)
{
int u = que[i];
for(int j = 0; j < v[u].size(); ++j)
{
if(--ind[v[u][j]] == 0) que[qt++] = v[u][j];
}
}
return qt == 26;
}
int main()
{
int n;
int flag=0;//儲存是否有字首完全相同一長一短不滿足字典序的情況
scanf("%d",&n);
for(int i=1;i<=n;i++) scanf("%s",str[i]);
for(int i=1;i<=n;i++)
{
for(int j=i+1;j<=n;j++)
{
int ff=0;
int len1=strlen(str[i]);
int len2=strlen(str[j]);
for(int pos1=0,pos2=0;pos1<len1&&pos2<len2;pos1++,pos2++)
{
if(str[i][pos1]!=str[j][pos2])
{
v[str[i][pos1]-'a'].push_back(str[j][pos2]-'a');
ind[str[j][pos2]-'a']++;
ff=1;
break;
}
}
if(ff==0&&len1>len2) flag=1;
}
}
if(topsort()&&flag==0)
{
for(int i=0;i<26;i++) printf("%c",(char)que[i]+'a');
}
else
{
printf("Impossible\n");
}
return 0;
}