1314-names for Babies
PDF (中文版) Statistics Forum
Time Limit:4 second (s) Memory limit:32 MB
Long ago, there was a strange kingdom. Peoples of different religions, different cultures used to live there. But as they were different, their names were also different. So, in schools, offices, it is quite tough to call someone using his/her name, because some names were too nounced by persons from different culture.
So, the king made a plan. He took a string s and integers p and q and made a rule that names of the babies should is a substring of s, and the L Ength should be between P and Q (inclusive).
Now is given S, p and Q you have to find the number of distinct names that can be made.
Input
Input starts with an integer T (≤100), denoting the number of test cases.
Each case is starts with a line containing a string S. The length of S'll be between 2 and 10000 (inclusive) and S contains lowercase 中文版 letters only. The next line contains the integer p and q (1≤p≤q≤length (S)).
Output
For each case, print the case number and the number of distinct names that can is made.
Sample Input
Output for Sample Input
1
ABCdef
2 5
Case 1:14
Note
This problem is used in contest 12. But the length of S is between 2 and 100.
Obvious suffix array problem, statistics of different substrings, but only when the statistics, only those lengths in the range of requirements of the string
/************************************************************************* > File Name:LightOJ1314.cpp > Author:alex > Mail: [email protected] > Created time:2015 April 09 Thursday 19:52 15 seconds *************************** *********************************************/#include <functional>#include <algorithm>#include <iostream>#include <fstream>#include <cstring>#include <cstdio>#include <cmath>#include <cstdlib>#include <queue>#include <stack>#include <map>#include <bitset>#include <set>#include <vector>using namespace STD;Const DoublePI =ACOs(-1.0);Const intINF =0x3f3f3f3f;Const DoubleEPS =1e-15;typedef Long LongLL;typedefPair <int,int> PLL;Charstr[12000];classsuffixarray{ Public:Static Const intN =12000;intInit[n];intX[n];intY[n];intRank[n];intSa[n];intHeight[n];intBuc[n];intLog[n];intdp[n][ -];intSizevoidClear () {size =0; }voidInsertintN) {init[size++] = n; }BOOLcmpint*r,intAintBintL) {return(R[a] = = R[b] && r[a + l] = = R[b + L]); }voidGetsa (intm = the)//m is generally the maximum value of +1{Init[size] =0;intL, p, *x = x, *y = y, n = size +1; for(inti =0; I < m; ++i) {Buc[i] =0; } for(inti =0; I < n; ++i) {++buc[x[i] = init[i]]; } for(inti =1; I < m; ++i) {Buc[i] + = buc[i-1]; } for(inti = n-1; I >=0; -i) {sa[--buc[x[i]] = i; } for(L =1, p =1; L <= N && p < n; m = p, l *=2) {p =0; for(inti = n-l; I < n; ++i) {y[p++] = i; } for(inti =0; I < n; ++i) {if(Sa[i] >= L) {y[p++] = sa[i]-l; } } for(inti =0; I < m; ++i) {Buc[i] =0; } for(inti =0; I < n; ++i) {++buc[x[y[i]]; } for(inti =1; I < m; ++i) {Buc[i] + = buc[i-1]; } for(inti = n-1; I >=0; -i) {sa[--buc[x[y[i]]] = y[i]; }intI for(Swap (x, y), x[sa[0]] =0, p =1, i =1; I < n; ++i) {X[sa[i]] = cmp (y, Sa[i-1], Sa[i], L)? P-1: p++; } } }voidGetHeight () {inth =0, n = size; for(inti =0; I <= N; ++i) {Rank[sa[i]] = i; } height[0] =0; for(inti =0; I < n; ++i) {if(H >0) {--h; }intJ =sa[rank[i]-1]; for(; i + H < n && j + H < n && init[i + h] = = Init[j + h]; ++h); Height[rank[i]-1] = h; } }//preprocessing logarithm of each number for RMQ, constant optimization voidInitlog () {log[0] = -1; for(inti =1; i < N; ++i) {Log[i] = (I & (i-1)) ? Log[i-1]: Log[i-1] +1; } }voidINITRMQ () {initlog ();intn = size;intLimit for(inti =0; I < n; ++i) {dp[i][0] = Height[i]; } for(intj =1; J <= Log[n]; ++J) {limit = (N-(1<< j)); for(inti =0; I <= limit; ++i) {Dp[i][j] = min (Dp[i][j-1], Dp[i + (1<< (J-1))][j-1]); } } }intLCP (intAintb) {intT A = Rank[a]; b = Rank[b];if(A > B) {Swap (A, b); }--b; t = log[b-a +1];returnMin (dp[a][t], dp[b-(1<< T) +1][t]); }voidSolveintPintQ) {intAns =0; for(inti =1; I <= size; ++i) {//Actual length range intr = Size-sa[i];intL = height[i-1] +1;if(q <= r && q >= L) {ans + = Q-max (l, p) +1; }Else if(q > R && P <= r) {ans + = R-max (l, p) +1; } }printf("%d\n", ans); }}sa;intMain () {intTintP, q;scanf("%d", &t);intIcase =1; while(t--) {sa.clear ();scanf('%s ', str);scanf("%d%d", &p, &q);intLen =strlen(str); for(inti =0; i < Len; ++i) {Sa.insert (Str[i]-' A '+1); } Sa.getsa ( -); Sa.getheight ();printf("Case%d:", icase++); Sa.solve (P, q); }return 0;}
LightOJ1314---Names for babies