D. Restoration of string
题意:给你n个字符串,让你构造一个终串,使得这n个字符串都是终串的最小频繁子串,如果不存在输出NO。 最频繁子串:出现次数最多的子串
怎么爬codeforces的数据,tags: 直接暴力怼??
#include<bits/stdc++.h> using namespace std; #pragma comment(linker, "/STACK:102400000,102400000") #define rep(i,a,b) for (int i=a; i<=b; ++i) #define per(i,b,a) for (int i=b; i>=a; --i) #define mes(a,b) memset(a,b,sizeof(a)) #define INF 0x3f3f3f3f #define MP make_pair #define PB push_back #define fi first #define se second typedef long long ll; const int N = 200005;int n, in[28]; char s[N]; bool vis[28], used[28]; vector< int > G[N]; void print(int x) {putchar('a'+x);vis[x] = true;if(G[x].size()==1)print(G[x][0]); } bool check(int x) {if(vis[x]) return false;vis[x]=true;if(G[x].size()==1)return check(G[x][0]);return true; } int main() {scanf("%d", &n);rep(i,1,n){scanf("%*c%s", s+1);mes(vis, false);for(int j=1; s[j]; ++j){int id = s[j]-'a';if(vis[id]) return 0*printf("NO\n");vis[id]=true, used[id]=true;if(j>1) {if(G[s[j-1]-'a'].size()==1 && G[s[j-1]-'a'][0]==s[j]-'a')continue;G[s[j-1]-'a'].PB(s[j]-'a');++in[s[j]-'a'];}if(G[s[j-1]-'a'].size()>1 || in[s[j]-'a']>1)return 0*printf("NO\n");}}mes(vis, false);rep(i,0,25)if(used[i]) {mes(vis, false);if(!check(i)) return 0*printf("NO\n");}mes(vis, false);rep(i,0,25)if(used[i] && !vis[i] && in[i]==0)print(i);puts("");return 0; }