Codeforces Round #291 (Div. 2) C. Watto and Mechanism [字典树]

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Watto, the owner of a spare parts store, has recently got an order for the mechanism that can process strings in a certain way. Initially the memory of the mechanism is filled with n strings. Then the mechanism should be able to process queries of the following type: "Given string s, determine if the memory of the mechanism contains string t that consists of the same number of characters as s and differs from s in exactly one position".

Watto has already compiled the mechanism, all that's left is to write a program for it and check it on the data consisting of n initial lines and m queries. He decided to entrust this job to you.

Input

The first line contains two non-negative numbers n and m (0 ≤ n ≤ 3·10⁵, 0 ≤ m ≤ 3·10⁵) — the number of the initial strings and the number of queries, respectively.

Next follow n non-empty strings that are uploaded to the memory of the mechanism.

Next follow m non-empty strings that are the queries to the mechanism.

The total length of lines in the input doesn't exceed 6·10⁵. Each line consists only of letters 'a', 'b', 'c'.

Output

For each query print on a single line "YES" (without the quotes), if the memory of the mechanism contains the required string, otherwise print "NO" (without the quotes).

Sample test(s)

Input

2 3
aaaaa
acacaca
aabaa
ccacacc
caaac

Output

YES
NO
NO

题意：n个串，m次查询，每次给一个字符串，问在原串中能不能找到一个串，与之长度相同，且只有一个字符不同。

题解：没什么好说的，字典树，强行有个数组不开成全局变量，T哭了，，，，

9856760	2015-02-15 11:39:10	njczy2010	C - Watto and Mechanism	GNU C++	Accepted	577 ms	99188 KB
9856712	2015-02-15 11:34:22	njczy2010	C - Watto and Mechanism	GNU C++	Time limit exceeded on test 32	3000 ms	101200 KB
9856707	2015-02-15 11:33:54	njczy2010	C - Watto and Mechanism	GNU C++	Memory limit exceeded on test 1	46 ms	262100 KB
9856697	2015-02-15 11:32:29	njczy2010	C - Watto and Mechanism	GNU C++	Time limit exceeded on test 32	3000 ms	90700 KB
9856633	2015-02-15 11:26:42	njczy2010	C - Watto and Mechanism	GNU C++	Time limit exceeded on test 32	3000 ms	89500 KB
9856553	2015-02-15 11:17:52	njczy2010	C - Watto and Mechanism	GNU C++	Wrong answer on test 4	15 ms	70700 KB
9856517	2015-02-15 11:14:26	njczy2010	C - Watto and Mechanism	GNU C++	Wrong answer on test 13	46 ms	70800 KB

  1 #include<iostream>
  2 #include<cstring>
  3 #include<cstdlib>
  4 #include<cstdio>
  5 #include<algorithm>
  6 #include<cmath>
  7 #include<queue>
  8 #include<map>
  9 #include<set>
 10 #include<stack>
 11 #include<string>
 12 
 13 #define N 300005
 14 #define M 1505
 15 //#define mod 10000007
 16 //#define p 10000007
 17 #define mod2 1000000000
 18 #define ll long long
 19 #define LL long long
 20 #define eps 1e-6
 21 //#define inf 2147483647
 22 #define maxi(a,b) (a)>(b)? (a) : (b)
 23 #define mini(a,b) (a)<(b)? (a) : (b)
 24 
 25 using namespace std;
 26 
 27 int n,m;
 28 int fff[10*N];
 29 
 30 typedef struct
 31 {
 32     char v;
 33     int mp[4];
 34 }PP;
 35 
 36 int tot;
 37 int cnt[10*N];
 38 
 39 PP p[10*N];
 40 char s[10*N];
 41 
 42 void insert(int l)
 43 {
 44     int i;
 45     int now=0;
 46     for(i=0;i<l;i++){
 47         if(p[now].mp[ s[i]-'a' ]==0){
 48             tot++;
 49             p[now].mp[ s[i]-'a' ]=tot;
 50             p[tot].v=s[i];
 51             memset(p[tot].mp,0,sizeof(p[tot].mp));
 52             now=tot;
 53         }
 54         else{
 55             now=p[now].mp[ s[i]-'a' ];
 56         }
 57     }
 58     cnt[now]++;
 59 }
 60 
 61 void ini()
 62 {
 63     memset(fff,0,sizeof(fff));
 64     int i;
 65     int l;
 66     tot=0;
 67     p[0].v='z';
 68     memset(p[0].mp,0,sizeof(p[0].mp));
 69 
 70     for(i=1;i<=n;i++){
 71         scanf("%s",s);
 72         l=strlen(s);
 73         fff[l]=1;
 74         insert(l);
 75     }
 76     //for(i=0;i<=tot;i++){
 77    //     printf(" i=%d v=%c cnt=%d\n",i,p[i].v,cnt[i]);
 78    // }
 79 }
 80 
 81 int check(int l,int cou,int now,int f)
 82 {
 83    // printf(" l=%d cou=%d now=%d v=%c cnt=%d f=%d\n",l,cou,now,p[now].v,cnt[now],f);
 84     if(cou==l){
 85         if(cnt[now]==0) return 0;
 86         if(f==1) return 1;
 87         else return 0;
 88     }
 89     if(f>=2) return 0;
 90     int i;
 91     int ff;
 92     for(i=0;i<=2;i++){
 93         if(p[now].mp[i]!=0){
 94             if(s[cou]==i+'a'){
 95                 ff=check(l,cou+1,p[now].mp[i],f);
 96             }
 97             else{
 98                 ff=check(l,cou+1,p[now].mp[i],f+1);
 99             }
100             if(ff==1) return 1;
101         }
102     }
103     return 0;
104 }
105 
106 void solve()
107 {
108     int i;
109     int l;
110     int flag;
111     for(i=1;i<=m;i++){
112         scanf("%s",s);
113         l=strlen(s);
114        // printf("l=%d\n",l);
115         if(fff[l]==0){
116             flag=0;
117         }
118         else{
119             flag=check(l,0,0,0);
120         }
121         if(flag==1){
122             printf("YES\n");
123         }
124         else{
125             printf("NO\n");
126         }
127     }
128 }
129 
130 void out()
131 {
132 
133 }
134 
135 int main()
136 {
137     //freopen("data.in","r",stdin);
138     //freopen("data.out","w",stdout);
139     //scanf("%d",&T);
140     //for(int ccnt=1;ccnt<=T;ccnt++)
141     //while(T--)
142     //scanf("%d%d",&n,&m);
143     while(scanf("%d%d",&n,&m)!=EOF)
144     {
145         ini();
146         solve();
147         out();
148     }
149     return 0;
150 }

posted on 2015-02-15 16:44 njczy2010 阅读(426) 评论(0) 编辑收藏举报