ACdream 1142 String opearation

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String opearation

Time Limit: 4000/2000MS (Java/Others)Memory Limit: 128000/64000KB (Java/Others)

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Problem Description

Give you a string S whose length is no more than 1000, and two operations.

Operation 1：read a string Str1，and add to S。S becomes S+str1

Operation 2：read a string Str2，and tell me whether it is a subsequence of S（not necessarily consecutive）

or the subsequence of S'，S' equals to S's shifting left（"bca"、"cab" all equals to "abc"'s shifting left)

Input

The first line is an integer T(1<=T<=150) ，the number of test cases;

For each test case，

The first line is a string S(length <= 1000)

The second line is a positive integer M (1<=M<=100)

The follow M lines:

Command string (Command is 1 or 2，and the string's length <= 100)

All the strings only consist of lowercase letters

Output

When operation 2 occurs，output "YES" if exists，otherwise "NO"

Sample Input

1abc31  d2  dca2  dbc

Sample Output

NOYES

Hint

if S = "abc", STR = "ef", then S + STR is "abcef", also the strcat funtion in C

Source

dream

Manager

admin

第一次做ACdream的手速赛，这道题虽然不难，但整个解题思路还是挺有意思的，所以就记录下来了。

第一次没看清，以为是要连续的，而且主字符串也是个环，WA了两发。后来发现错误后，很荣欣地AC了。

因为最多只有10000左右的字符，我是记录了每一个字符的位置，用vector记录，然后查找是否是子串时，子串成环，每一种可能都搜一遍，从第一个字符开始搜，每搜一个字符记录这个字符搜到的位置，下一次搜要大于这个位置，至于搜，当然是二分啦。。。

代码：

#include<iostream>#include<cstdio>#include<cstdlib>#include<ctime>#include<string>#include<cstring>#include<algorithm>#include<queue>#include<stack>#include<vector>#include<cmath>#define rep(i,n) for(i=1;i<=n;i++)#define MM(a,t) memset(a,t,sizeof(a))#define INF 1e9typedef long long ll;#define mod 1000000007using namespace std;char st[12000];char sst[120];int n,l;vector<int> ki[30];void songchi(int ii,int l2){    int i,j;    for(i=ii;i<l2;i++)    {        int index=int(st[i]-'a');        ki[index].push_back(i);    }}void fun1(){  int i,j,ii=l;  l+=strlen(sst);  for(i=ii;i<l;i++)    st[i]=sst[i-ii];  songchi(ii,l);   }int so(int index,int pr){    int l,r,res=-1;    l=0; r=ki[index].size()-1;    while(l<=r){        int mid=(l+r)/2;        if(ki[index][mid]>=pr){          res=ki[index][mid];          r=mid-1;         }        else l=mid+1;    }    if(res==-1) return -1;    else        return res;}bool fun2(){    int i,j,l2=strlen(sst);         if(l2>l) return false;    if(l2==1){        if(ki[int(sst[0]-'a')].size()!=0) return true;        else                              return false;    }    for(i=0;i<l2;i++)    {      int s2=i,index=int(sst[i]-'a');      int pr=so(index,0);      if(pr==-1) return false;      else{        s2=(s2+1)%l2;        bool ff=true;        while(s2!=i){            index=int(sst[s2]-'a');            pr=so(index,pr+1);            if(pr==-1){                ff=false;                break;            }            s2=(s2+1)%l2;          }        if(ff) return true;      }    }    return false;}int main(){    int i,j,T;         scanf("%d",&T);    while(T--){        for(i=0;i<26;i++) ki[i].clear();        scanf("%s",&st);        l=strlen(st);        songchi(0,l);        scanf("%d",&n);        rep(i,n){            int op;            scanf("%d",&op);            scanf("%s",&sst);            if(op==1) fun1();            else {              if(fun2()) printf("YES\n");              else       printf("NO\n");            }        }    }          return 0;}

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