hdu 3973 字符串hash+線段樹

 

Problem Description You are given some words {Wi}. Then our stupid AC will give you a very long string S. AC is stupid and always wants to know whether one substring from S exists in the given words {Wi} .

For example, S = abcd, and the given words {Wi} = {bc, ad, dd}. Then Only S[2..3] = bc exists in the given words. (In this problem, the first element of S has the index 0.)

However, this is toooooooooooo easy for acmers ! The stupid and evil AC will now change some letters in S. So could you solve this problem now?
Input The first line is one integer T indicates the number of the test cases. (T <=20)

Then for every case, there is one integer n in the first line indicates the number of the given words(The size of the {Wi}) . Then n lines has one string which only has 'a'- 'z'. (1 <= n <= 10000, sigma|Wi| <= 2000000) .

Then one line has one string S, here |S| <= 100000.

Then one integer m, indicating the number of operations. (1 <= m <= 100000)

Then m lines , each line is the operation:

(1)Q L R , tell AC whether the S[L..R] exists in the given strings ;

(2)C X Y , chang S[X] to Y, here Y : 'a'-'z' .www.2cto.com
Output First output “Case #idx:” in a single line, here idx is the case number count from 1.Then for each Q operation, output Yes if S[L..R] exists in the given strings, otherwise output No.
Sample Input

1
2
ab 
ioc 
ipcad 
6 
Q 0 2 
Q 3 4 
C 1 o 
C 4 b 
Q 0 2 
Q 3 4

Sample Output

Case #1:
No
No 
Yes 
Yes

/**
hdu 3973  線段樹單點更新區間求值+字符串hash
題目大意：給定多個字符串，然後給定一個大串，對該串進行單點更新和區間查詢，查詢的區間子串是不是在已知的字符串中出現
解題思路：對字符串進行hash處理采用線段樹來進行更新，用set存放字符串的哈希值。至於怎麼哈希和大白書上的思路差不多只是這裡是表示的前綴
*/
#include 
#include 
#include 
#include 
using namespace std;

const int maxn=100010;
const int seed=31;
typedef unsigned long long LL;

struct note
{
    int l,r;
    LL hashes;
}tree[maxn*4];
char str[2000100];
LL Hash[maxn];
int n;

void init()
{
    Hash[0]=1;
    for(int i=1;i>1;
    if(l<=mid) update(l,root<<1);
    else  update(l,root<<1|1);
    tree[root].hashes=tree[root<<1].hashes*Hash[tree[root].r-mid]+tree[root<<1|1].hashes;
}

LL query(int l,int r,int root)
{
  //  printf(**
);
    if(tree[root].l==l&&tree[root].r==r)
        return tree[root].hashes;
    int mid=(tree[root].r+tree[root].l)>>1;
    if(r<=mid)return query(l,r,root<<1);
    else if(l>mid)return query(l,r,root<<1|1);
    return query(l,mid,root<<1)*Hash[r-mid]+query(mid+1,r,root<<1|1);
}
int main()
{
    int T,tt=0;
    init();
    scanf(%d,&T);
    while(T--)
    {
        printf(Case #%d:
,++tt);
        scanf(%d,&n);
        setmp;
        for(int i=0;i