Codeforces 510C
题意:给出n个名字,问是否存在一个字典序(这个字典序是任意的)使得这些名字是按这个字典序输出的,如果存在输出任意一个可行的字典序
思路:拓扑排序,关键是建图,第i个和i+1个建一条由i指向i+1的单向边,建图是a-z26个字母间建图,每2个相邻的单词之间排序的依据是第一个不相同的字母,由这个不相同的字母之间建边,如果没有不相同的则不建边,这里有一个坑,就是 aaa aaaa这种数据,要特判
AC代码:
#include "iostream" #include "string.h" #include "stack" #include "queue" #include "string" #include "vector" #include "set" #include "map" #include "algorithm" #include "stdio.h" #include "math.h" #pragma comment(linker, "/STACK:102400000,102400000") #define ll long long #define endl ("\n") #define bug(x) cout<<x<<" "<<"UUUUU"<<endl; #define mem(a,x) memset(a,x,sizeof(a)) #define mp(x,y) make_pair(x,y) #define pb(x) push_back(x) #define ft (frist) #define sd (second) #define lrt (rt<<1) #define rrt (rt<<1|1) using namespace std; const long long INF = 1e18+1LL; const int inf = 1e9+1e8; const int N=1e5+100; const ll mod=1e9+7; struct AN{ int id,num; bool friend operator< (AN a, AN b){ return a.num<b.num; } }ans[160]; char s[155][155]; int n,head[155],nex[305],to[305],tot=1,in[300],cnt; bool vis[155]; void add(int u, int v){ to[tot]=v; nex[tot]=head[u]; head[u]=tot++; } void topu(int u){ queue<int> Q; while(!Q.empty()) Q.pop(); Q.push(u); vis[u]=1; while(!Q.empty()){ int now=Q.front(); Q.pop(); ans[now].num=++cnt; for(int i=head[now]; i!=-1; i=nex[i]){ int v=to[i]; in[v]--; if(in[v]==0 && vis[v]==0){ Q.push(v); vis[v]=1; } } } } int main(){ ios::sync_with_stdio(false),cin.tie(0),cout.tie(0); cin>>n; mem(head,-1); for(int _=1; _<=n; ++_){ cin>>s[_]; } for(int i=1; i<n; ++i){ int k=0; int l1=strlen(s[i]), l2=strlen(s[i+1]); while(s[i][k]==s[i+1][k] && k<=l1 && k<=l2){ ++k; } if(k>=l1 || k>=l2){ if(l2<l1){ cout<<"Impossible\n"; return 0; } continue; } add(s[i][k]-'a'+1, s[i+1][k]-'a'+1); in[s[i+1][k]-'a'+1]++; } for(int i=1; i<=26; ++i){ //cout<<in[i]<<" "; if(in[i]==0 && vis[i]==0) topu(i);//bug(ans[i].num) } for(int i=1; i<=26; ++i){ if(ans[i].num==0){ cout<<"Impossible\n"; return 0; } } for(int i=1; i<=26; ++i) ans[i].id=i; sort(ans+1,ans+1+26); for(int i=1; i<=26; ++i) cout<<char(ans[i].id+'a'-1); return 0; }