HDU 3973 字符串HASH

题目链接

题意：

给定n个字符串和一个字符串S有m次查询，查询分两种：

Q L R 询问[L,R]区间所形成的子串是不是n个字符串中的一个，是输出YES，否则输出NO

C X Y 将位置X上的字符修改成Y

思路：

将n个字符串HASH处理后进行排序，将字符串以线段树方式进行处理

Q L R操作：将[L,R]区间所形成的子串的HASH值得出，在n个字符串所形成的数组中进行二分

C X Y操作：线段树更新操作

C++代码：

#include<bits/stdc++.h>
using namespace std;
typedef unsigned long long ull;
const ull maxn = 100010;
const ull mod1 = 1e9+7;
const ull mod2 = 1e9+9;
const ull base1 = 131;
const ull base2 = 233;

struct node
{
    ull a,b;
    friend bool operator<( const node&a , const node&b )
    {
        if ( a.a==b.a )
            return a.b<b.b;
        else
            return a.a<b.a;
    }
}data[10010];

ull p1[maxn],p2[maxn];

ull GetHash1( char s[] )
{
    int len = strlen(s+1);
    ull res = 0;
    for ( int i=1 ; i<=len ; i++ )
        res = (res*base1+(ull)s[i])%mod1;
    return res;
}

ull GetHash2( char s[] )
{
    int len = strlen(s+1);
    ull res = 0;
    for ( int i=1 ; i<=len ; i++ )
        res = (res*base2+(ull)s[i])%mod2;
    return res;
}

char s[maxn];

int len;
ull sum1[maxn<<2];
ull sum2[maxn<<2];

void build( int l , int r , int rt )
{
    int lson = rt<<1;
    int rson = rt<<1|1;
    int mid = ( l+r )>>1;
    if ( l==r )
    {
        sum1[rt] = (ull)s[l];
        sum2[rt] = (ull)s[l];
        return;
    }
    build ( l , mid , lson );
    build ( mid+1 , r , rson );
    sum1[rt] = (sum1[lson]*p1[r-mid]+sum1[rson])%mod1;
    sum2[rt] = (sum2[lson]*p2[r-mid]+sum2[rson])%mod2;
}

void updata( int l , int r , int rt , int pos , char ch )
{
    if ( l==pos&&r==pos )
    {
        sum1[rt] = (ull)ch;
        sum2[rt] = (ull)ch;
        return;
    }
    int lson = rt<<1;
    int rson = rt<<1|1;
    int mid = ( l+r )>>1;
    if ( pos<=mid )
        updata ( l , mid , lson , pos , ch );
    else
        updata ( mid+1 , r , rson , pos , ch );
    sum1[rt] = (sum1[lson]*p1[r-mid]+sum1[rson])%mod1;
    sum2[rt] = (sum2[lson]*p2[r-mid]+sum2[rson])%mod2;
}

ull query1( int l , int r , int rt , int L , int R )
{
    if ( l==L&&r==R ) return sum1[rt];
    int lson = rt<<1;
    int rson = rt<<1|1;
    int mid = ( l+r )>>1;
    if ( R<=mid )
        return query1( l , mid , lson , L , R );
    else if ( L>mid )
        return query1( mid+1 , r , rson , L , R );
    else
        return ( query1( l , mid , lson , L , mid )*p1[R-mid]%mod1+query1( mid+1 , r , rson , mid+1 , R ) )%mod1;
}

ull query2( int l , int r , int rt , int L , int R )
{
    if ( l==L&&r==R ) return sum2[rt];
    int lson = rt<<1;
    int rson = rt<<1|1;
    int mid = ( l+r )>>1;
    if ( R<=mid )
        return query2( l , mid , lson , L , R );
    else if ( L>mid )
        return query2( mid+1 , r , rson , L , R );
    else
        return ( query2( l , mid , lson , L , mid )*p2[R-mid]%mod2+query2( mid+1 , r , rson , mid+1 , R ) )%mod2;
}

int main()
{
    p1[0] = 1;
    p2[0] = 1;
    for ( int i=1 ; i<maxn ; i++ )
    {
        p1[i] = p1[i-1]*base1%mod1;
        p2[i] = p2[i-1]*base2%mod2;
    }
    int T,cas=0; scanf ( "%d" , &T );
    while ( T-- )
    {
        printf ( "Case #%d:\n" , ++cas );
        int n; scanf ( "%d" , &n );
        for ( int i=1 ; i<=n ; i++ )
        {
            scanf ( "%s" , s+1 );
            data[i].a = GetHash1(s);
            data[i].b = GetHash2(s);
        }
        sort ( data+1 , data+n+1 );
        scanf ( "%s" , s+1 );
        len = strlen(s+1);
        build ( 1 , len , 1 );
        int m; scanf ( "%d" , &m );
        for ( int i=1 ; i<=m ; i++ )
        {
            char ch[5],C[5];
            int x,y;
            scanf ( "%s" , ch );
            if ( ch[0]=='Q' )
            {
                scanf ( "%d%d" , &x , &y );
                ull t1 = query1( 1 , len , 1 , x+1 , y+1 );
                ull t2 = query2( 1 , len , 1 , x+1 , y+1 );
                int l=1,r=n,mid,ans=-1;
                while( l<=r )
                {
                    mid = (l+r)>>1;
                    if ( data[mid].a==t1&&data[mid].b==t2 )
                    {
                        ans = mid;
                        break;
                    }
                    else
                    {
                        if ( data[mid].a>t1||(data[mid].a==t1&&data[mid].b>t2) )
                            r = mid-1;
                        else
                            l = mid+1;
                    }
                }
                if ( ans==-1 )
                    printf ( "No\n" );
                else
                    printf ( "Yes\n" );
            }
            else
            {
                scanf ( "%d%s" , &x , C );
                updata( 1 , len , 1 , x+1 , C[0] );
            }
        }
    }
    return 0;
}