标签：tar code sub lse tac and term else long
Give a string SS and NN string T_iTi? , determine whether T_iTi? is a subsequence of SS.
If ti is subsequence of SS, print YES,else print NO.
If there is an array \lbrace K_1, K_2, K_3,\cdots, K_m \rbrace{K1?,K2?,K3?,?,Km?} so that 1 \le K_1 < K_2 < K_3 < \cdots < K_m \le N1≤K1?<K2?<K3?<?<Km?≤N and S_{k_i} = T_iSki??=Ti?, (1 \le i \le m)(1≤i≤m), then T_iTi? is a subsequence of SS.
Input
The first line is one string SS,length(SS) \le 100000≤100000
The second line is one positive integer N,N \le 100000N,N≤100000
Then next nn lines,every line is a string T_iTi?, length(T_iTi?) \le 1000≤1000
Output
Print NN lines. If the ii-th T_iTi? is subsequence of SS, print YES, else print NO.
样例输入复制
abcdefg
3
abc
adg
cba
样例输出复制
YES
YES
NO
#include<iostream> #include<cstdio> #include<algorithm> #include<string> #include<cstring> #include<queue> #include<stack> #include<cmath> #include<set> #include<map> using namespace std; #define ll long long #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 typedef pair<int,int>P; const int INF=0x3f3f3f3f; const int N=1000010,mod=32767; char s[N],p[N]; int nex[N][30],now[N]; //nex[i][j]:=表示第i个字符后面第一次出现字符j（a-z用0-25表示）的位置。 //我们从后往前求，now[j]:=字符j从后往前数最晚出现的位置（now数组初始化为-1） void init() { memset(now,-1,sizeof(now)); int len=strlen(s); for(int i=len-1; i>=0; i--) { for(int j=0; j<26; j++) { nex[i][j]=now[j]; } now[s[i]-‘a‘]=i; } } int main() { int n,len,loc,flag; scanf("%s",s); init(); scanf("%d",&n); for(int i=1; i<=n; i++) { scanf("%s",p); len=strlen(p); loc=now[p[0]-‘a‘]; if(loc==-1) printf("NO\n"); else { flag=0; for(int i=1; i<len; i++) { loc=nex[loc][p[i]-‘a‘]; if(loc==-1) { flag=1; break; } } if(!flag) printf("YES\n"); else printf("NO\n"); } } }
标签：tar code sub lse tac and term else long
原文地址：https://www.cnblogs.com/Agnel-Cynthia/p/10947280.html
