HDU 3973 AC ' s string string hash

HDU 3973

By converting a string into an integer through a hash function, it is possible to make the hash value virtually non-conflicting (this is the key point, almost no = =!). , you can also consider the process of hashing conflicts, but this problem is not necessary, and then use the segment tree to maintain the modified hash value.

Because the input string is only 26, consider using a prime number greater than or equal to 26 as the binary, then convert the original string into a P-binary mod 2^63 (also equivalent to a natural overflow), then this number is stored in the map, and then use the segment tree to maintain the hash value of the long range, Hash[l, R] Represents the result of converting a string of intervals [L, R] to a P-binary number (hash procedure), then for the two sub-bands [L, M] and [M + 1, R] of the current interval [L, R]:

Hash[l, R] = Hash[l, M] + hash[m + 1, r] * (p ^ (m-l + 1))

Using the Segment tree dynamic maintenance, query interval hash value, using map lookup, then the problem is solved.

1 //#pragma COMMENT (linker, "/stack:1677721600")2#include <map>3#include <Set>4#include <stack>5#include <queue>6#include <cmath>7#include <ctime>8#include <vector>9#include <cstdio>Ten#include <cctype> One#include <cstring> A#include <cstdlib> -#include <iostream> -#include <algorithm> the using namespacestd; - #defineINF 0x3f3f3f3f - #defineInf (-((LL) 1<<40)) - #defineLson k<<1, L, (L + R) >>1 + #defineRson k<<1|1, ((L + R) >>1) + 1, R - #defineMem0 (a) memset (A,0,sizeof (a)) + #defineMem1 (a) memset (A,-1,sizeof (a)) A #defineMem (A, B) memset (A, B, sizeof (a)) at #defineFIN freopen ("In.txt", "R", stdin) - #defineFOUT freopen ("OUT.txt", "w", stdout) - #defineRep (I, A, b) for (int i = A; I <= B; i + +) - #defineDec (i, A, b) for (int i = A; I >= b; i-) -  -template<classT> T MAX (t A, T b) {returna > B?a:b;} intemplate<classT> T MIN (t A, T b) {returnA < b?a:b;} -template<classT> T GCD (t A, T b) {returnB? GCD (b, a%b): A; } totemplate<classT> T LCM (t A, T b) {returnA/GCD (A, b) *b; } +  - //typedef __int64 LL; thetypedefLong LongLL; * Const intMAXN =100000+ -; $ Const intMAXM =110000;Panax Notoginseng Const DoubleEPS = 1e-8; -LL MOD =1000000007; the  + ConstLL P = -; A  the intT, N, m, cas =0, L, R; + LL MI[MAXN]; - Charstr[2000001], ch; $Map<ll,int>MP; $  - voidinit () { -mi[0] =1; theRep (I,1, MAXN-1) { -Mi[i] = mi[i-1] *P;Wuyi     } the } -  Wu intTonum (Charch) { -     returnCH-'a'+1; About } $  - LL Hash () { -LL ans =0; -     intLen =strlen (str); ADec (i, Len-1,0) { +Ans = ans * P +Tonum (Str[i]); the     } -     returnans; $ } the  the structSegtree { theLL MA[MAXN <<2]; the  -     voidUpdateintKintLintRintPintv) { in         if(L = = R) {Ma[k] = V;return ; } the         intMid = (L + R) >>1; the         if(Mid >=p) Update (Lson, p, v); About         ElseUpdate (Rson, p, v); theMa[k] = ma[k <<1] + ma[k <<1|1] * mi[mid-l +1]; the     } the  +LL Query (intKintLintRintLintr) { -         if(R < L | | R < L)return 0; the         if(l <= L && R <= R)returnMa[k];BayiLL le = Query (Lson, L, r), RI =query (Rson, L, R); the         intMid = (L + R) >>1; the         returnLe + ri * Mi[max (Mid-max (L, L) +1,0)]; -     } -  the }st; the  the intMain () the { - //FIN; the init (); theCIN >>T; the      while(t--) {94 mp.clear (); the mem0 (st.ma); thescanf"%d%*c", &n); theRep (I,0N1) {98scanf"%s", str); AboutMp[hash ()] =1; -         }101scanf"%s", str);102         intLen =strlen (str);103Rep (I,0, Len-1) {104St.update (1,1, Len, i +1, Tonum (Str[i])); the         }106printf"Case #%d:\n", ++CAs);107scanf"%d%*c", &m);108          while(M--) {109scanf"%c", &ch); the             if(ch = ='Q') {111scanf"%d%d%*c", &l, &R); theprintf"%s\n", Mp[st.query (1,1, Len, ++l, ++r)]?"Yes":"No");113             } the             Else { thescanf"%d%c%*c", &l, &ch); theSt.update (1,1, Len, + +l, Tonum (ch));117             }118         }119     } -     return 0;121}

HDU 3973 AC ' s string string hash